Summer is a great time to explore cool problems to keep your skills sharp!  Schedule a class today!

G
Topic
First Poster
Last Poster
k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Inspired by qrxz17
sqing   5
N 2 minutes ago by pooh123
Source: Own
Let $a, b,c>0 ,(a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) = 27 $. Prove that $$a+b+c\geq 3\sqrt {3}$$
5 replies
sqing
4 hours ago
pooh123
2 minutes ago
Basic ideas in junior diophantine equations
Maths_VC   1
N 9 minutes ago by grupyorum
Source: Serbia JBMO TST 2025, Problem 3
Determine all positive integers $a, b$ and $c$ such that
$2$ $\cdot$ $10^a + 5^b = 2025^c$
1 reply
Maths_VC
Tuesday at 7:54 PM
grupyorum
9 minutes ago
Inspired by qrxz17
sqing   3
N 20 minutes ago by sqing
Source: Own
Let $ a,b,c $ be reals such that $ (a^2+b^2)^2 + (b^2+c^2)^2 +(c^2+a^2)^2 = 28 $ and $  (a^2+b^2+c^2)^2 =16. $ Find the value of $ a^2(a^2-1) + b^2(b^2-1)+c^2(c^2-1).$
3 replies
sqing
4 hours ago
sqing
20 minutes ago
Hardest in ARO 2008
discredit   30
N 22 minutes ago by Phat_23000245
Source: ARO 2008, Problem 11.8
In a chess tournament $ 2n+3$ players take part. Every two play exactly one match. The schedule is such that no two matches are played at the same time, and each player, after taking part in a match, is free in at least $ n$ next (consecutive) matches. Prove that one of the players who play in the opening match will also play in the closing match.
30 replies
discredit
Jun 11, 2008
Phat_23000245
22 minutes ago
Find the value
sqing   12
N 25 minutes ago by Phat_23000245
Source: 2024 China Fujian High School Mathematics Competition
Let $f(x)=a_6x^6+a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0,$ $a_i\in\{-1,1\} ,i=0,1,2,\cdots,6 $ and $f(2)=-53 .$ Find the value of $f(1).$
12 replies
sqing
Jun 22, 2024
Phat_23000245
25 minutes ago
Midpoints of arcs form a similar triangle
v_Enhance   19
N 25 minutes ago by Adywastaken
Source: USA TSTST 2013, Problem 1
Let $ABC$ be a triangle and $D$, $E$, $F$ be the midpoints of arcs $BC$, $CA$, $AB$ on the circumcircle. Line $\ell_a$ passes through the feet of the perpendiculars from $A$ to $DB$ and $DC$. Line $m_a$ passes through the feet of the perpendiculars from $D$ to $AB$ and $AC$. Let $A_1$ denote the intersection of lines $\ell_a$ and $m_a$. Define points $B_1$ and $C_1$ similarly. Prove that triangle $DEF$ and $A_1B_1C_1$ are similar to each other.
19 replies
v_Enhance
Aug 13, 2013
Adywastaken
25 minutes ago
find question
mathematical-forest   4
N 26 minutes ago by GreekIdiot
Are there any contest questions that seem simple but are actually difficult? :-D
4 replies
mathematical-forest
2 hours ago
GreekIdiot
26 minutes ago
4 var inequality
SunnyEvan   1
N 28 minutes ago by SunnyEvan
Source: Own
Let $ x,y,z,t \in R^+ ,$ such that : $ (x+y+z+t)^2 = x+y+z+t + (x+z)(y+t) $ and $ x \geq y \geq z \geq t .$
Try to prove or disprove : $$ \frac{2 \sqrt{x+y+z+t +(x+t)(y+z)}}{x^2+y^2+z^2+t^2 +3xz+3yt+xt+yz} \geq \frac{11(x+y)(z+t)-(x+y+z+t)}{x+y+z+t +(x+z)(y+t)} $$
1 reply
SunnyEvan
5 hours ago
SunnyEvan
28 minutes ago
Serbian selection contest for the IMO 2025 - P6
OgnjenTesic   14
N an hour ago by math90
Source: Serbian selection contest for the IMO 2025
For an $n \times n$ table filled with natural numbers, we say it is a divisor table if:
- the numbers in the $i$-th row are exactly all the divisors of some natural number $r_i$,
- the numbers in the $j$-th column are exactly all the divisors of some natural number $c_j$,
- $r_i \ne r_j$ for every $i \ne j$.

A prime number $p$ is given. Determine the smallest natural number $n$, divisible by $p$, such that there exists an $n \times n$ divisor table, or prove that such $n$ does not exist.

Proposed by Pavle Martinović
14 replies
OgnjenTesic
May 22, 2025
math90
an hour ago
Three Distinct Divisors Sum to 2022
ike.chen   31
N an hour ago by Jupiterballs
Source: ISL 2022/N1
A number is called Norwegian if it has three distinct positive divisors whose sum is equal to $2022$. Determine the smallest Norwegian number.
(Note: The total number of positive divisors of a Norwegian number is allowed to be larger than $3$.)
31 replies
ike.chen
Jul 9, 2023
Jupiterballs
an hour ago
MOP 2012 Inequality
holdmyquadrilateral   3
N an hour ago by Adywastaken
Source: MOP 2012
For $a,b,c>0$, prove that
\[\frac{a^3}{(b-c)^2+bc}+\frac{b^3}{(c-a)^2+ca}+\frac{c^3}{(a-b)^2+ab}\ge a+b+c.\]
3 replies
holdmyquadrilateral
Mar 11, 2023
Adywastaken
an hour ago
strange geometry problem
Zavyk09   2
N 2 hours ago by Zavyk09
Source: own
Let $ABC$ be a triangle with circumcenter $O$ and internal bisector $AD$. Let $AD$ cuts $(O)$ again at $M$ and $MO$ cuts $(O)$ again at $N$. Point $L$ lie on $AD$ such that $(AD, LM) = -1$. The line pass through $L$ and perpendicular to $AD$ intersects $NC, NB$ at $P, Q$ respectively. Let circumcircle of $\triangle NPQ$ cuts $(O)$ at $G \ne N$. Prove that $\angle AGD = 90^{\circ}$.
2 replies
Zavyk09
Yesterday at 4:32 PM
Zavyk09
2 hours ago
exponential diophantine with factorials
skellyrah   5
N 2 hours ago by skellyrah
find all non negative integers (x,y) such that $$ x! + y! = 2025^x + xy$$
5 replies
skellyrah
Feb 24, 2025
skellyrah
2 hours ago
A second final attempt to make a combinatorics problem
JARP091   2
N 2 hours ago by JARP091
Source: At the time of writing this problem I do not know the source if any
Arthur Morgan is playing a game.

He has $n$ eggs, each with a hardness value $k_1, k_2, \dots, k_n$, where $\{k_1, k_2, \dots, k_n\}$ is a permutation of the set $\{1, 2, \dots, n\}$. He is throwing the eggs from an $m$-floor building.

When the $i$-th egg is dropped from the $j$-th floor, its new hardness becomes
\[
\left\lfloor \frac{k_i}{j+1} \right\rfloor.
\]If $\left\lfloor \frac{k_i}{j+1} \right\rfloor = 0$, then the egg breaks and cannot be used again.

Arthur can drop each egg from a particular floor at most once.
For which values of $n$ and $m$ can Arthur always determine the correct ordering of the eggs according to their initial hardness values?
Note: The problem might be wrong or too easy
2 replies
JARP091
May 25, 2025
JARP091
2 hours ago
Infimum of decreasing sequence b_n/n^2
a1267ab   35
N May 6, 2025 by shendrew7
Source: USA Winter TST for IMO 2020, Problem 1 and TST for EGMO 2020, Problem 3, by Carl Schildkraut and Milan Haiman
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?

Carl Schildkraut and Milan Haiman
35 replies
a1267ab
Dec 16, 2019
shendrew7
May 6, 2025
Infimum of decreasing sequence b_n/n^2
G H J
G H BBookmark kLocked kLocked NReply
Source: USA Winter TST for IMO 2020, Problem 1 and TST for EGMO 2020, Problem 3, by Carl Schildkraut and Milan Haiman
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
a1267ab
223 posts
#1 • 5 Y
Y by centslordm, Pluto1708, megarnie, Adventure10, kub-inst
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?

Carl Schildkraut and Milan Haiman
This post has been edited 3 times. Last edited by a1267ab, Dec 16, 2019, 6:11 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
TheUltimate123
1740 posts
#2 • 9 Y
Y by Idio-logy, Williamgolly, Doxuantrong, ashrith9sagar_1, CALCMAN, centslordm, Pluto1708, Adventure10, MS_asdfgzxcvb
The answer is $0\le r\le1/2$.
Claim. $r=1/2$ works, and is maximal.

Proof. To achieve $r=1/2$, take $b_n=n(n+1)/2$, from which \[\frac{b_n}{n^2}=\frac{n(n+1)}{2n^2}=\frac{n+1}{2n}=\frac12+\frac1{2n},\]which clearly satisfies the problem condition. We inductively show that $b_n\le n(n+1)/2$. The base case has been given to us. Now, if the hypothesis holds for all integers less than $n$, then \[\frac{b_n}{n^2}<\frac{b_{n-1}}{(n-1)^2}\le\frac n{2(n-1)}\implies b_n<\frac{n^3}{2(n-1)}.\]It is easy to verify the largest possiblie $b_n$ is $n(n+1)/2$, as claimed. $\blacksquare$
Claim. All $r<1/2$ work.

Proof. Consider the sequence $(a_n)$ defined by $a_n:=\left\lceil kn^2\right\rceil+n$. Since $a_n$ is $O(n^2)$ and $k<1/2$, there exists $N$ such that for all $n\ge N$, $a_n/n^2<1/2$. I claim the sequence \[b_n:=\begin{cases}n(n+1)/2&\text{ for }n<N\\ a_n&\text{ for }n\ge N\end{cases}\]works. By definition of $N$, $b_n/n^2>b_{n+1}/(n+1)^2$ for $n<N$, so it suffices to verify $a_n/n^2$ is strictly decreasing for $n\ge N$.

In other words, we want to show that \[L:=\frac{\left\lceil kn^2\right\rceil+n}{n^2}>\frac{\left\lceil k(n+1)^2\right\rceil+n+1}{(n+1)^2}=:R\]for all $n\ge N$. Since $\left\lceil kn^2\right\rceil\ge kn^2$, \[L\ge\frac{kn^2+n}{n^2}=k+\frac1n,\]and similarly since $\left\lceil k(n+1)^2\right\rceil<k(n+1)^2+1$, \[R<\frac{k(n+1)^2+n+2}{(n+1)^2}=\frac1k+\frac{n+2}{(n+1)^2},\]so it suffices to verify that \[\frac1n\ge\frac{n+2}{(n+1)^2}\iff(n+1)^2\ge n(n+2),\]which is true. $\blacksquare$
Combining these two claims, we are done.
This post has been edited 1 time. Last edited by TheUltimate123, Dec 16, 2019, 5:49 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
spartacle
538 posts
#3 • 3 Y
Y by pad, centslordm, Adventure10
Sad... I essentially discovered this construction, but didn't think of replacing the initial "too large" terms with triangular numbers.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jeff10
1117 posts
#4 • 2 Y
Y by centslordm, Adventure10
Another Construction
This post has been edited 1 time. Last edited by jeff10, Dec 16, 2019, 8:20 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
MarkBcc168
1595 posts
#5 • 6 Y
Y by Mathphile01, SpecialBeing2017, MintTea, centslordm, Adventure10, Mango247
P1

The answer is $[0,\tfrac{1}{2}]$. We prove the bound first. Consider the following claim.

Claim: $b_n\leq\frac{n(n+1)}{2}$ for all positive integer $n$.

Proof: Induct on $n$. The base case $n=1$ is obvious. Assume that $b_{n-1}\leq\tfrac{n(n-1)}{2}$. We will prove that $b_{n}\leq\tfrac{n(n+1)}{2}$. Note that
\begin{align*}
b_{n} &< \frac{n^2}{(n-1)^2}\cdot\frac{n(n-1)}{2} \\
&= \frac{n^3}{2(n-1)} \\
&= \frac{n^3-n}{2(n-1)} + \frac{n}{2(n-1)} \\
&< \frac{n(n+1)}{2}+1
\end{align*}hence we are done.$\blacksquare$

The claim immediately implies the bound. Now we move on to the construction part.

The equality case above $b_n=\tfrac{n(n+1)}{2}$ gives $\tfrac{1}{2}$. Now we give a sequence with converge to $L$ for $0<L<\tfrac{1}{2}$. Define
$$s_n = \frac{1}{n^2} + \frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + \hdots$$Let $M$ be the smallest positive integer which $s_M<\tfrac{1}{2}-L$. We define the sequence $b_n$ by
$$b_n = \begin{cases}
\frac{n(n+1)}{2} & n<M^{2019} \\
\lfloor n^2(L+s_n)\rfloor & \text{otherwise.}
\end{cases}$$By the condition, for large $n$ we have
$$\frac{b_n}{n^2}\in \left(L+s_n+\frac{1}{n^2}, L+s_n\right] = (L+s_{n+1}, L+s_n).$$Intervals of this type are disjoint. This gives the strictly increasing. Moreover, $L<\tfrac{b_n}{n^2}\leq L+s_n$ thus the sequence $\tfrac{b_n}{n^2}$ converges to $L$ as desired.
This post has been edited 1 time. Last edited by MarkBcc168, Dec 17, 2019, 10:50 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IndoMathXdZ
694 posts
#6 • 4 Y
Y by FISHMJ25, MintTea, centslordm, Adventure10
I claim that any real numbers $0 \le r \le \frac{1}{2}$ satisfy this.

Notice that as $b_n \in \mathbb{N}$, then $\frac{b_n}{n^2} \in \mathbb{Q}^+$ for every $n \in \mathbb{N}$. This proves that $r \ge 0$.
To prove that $0$ is achievable, take $b_n = 1$ for all $n \in \mathbb{N}$ and we have
\[ \lim_{n \to \infty} \frac{b_n}{n^2} = \lim_{n \to \infty} \frac{1}{n^2} = 0\]To prove that $r \le \frac{1}{2}$ is maximum, notice that from the problem's constraint:
\[ \frac{b_{k+1}}{(k+1)^2} < \frac{b_k}{k^2} \]We'll prove by induction that $b_k \le \frac{k(k+1)}{2}$ for every $k \in \mathbb{N}$.
For $k = 1$, we have $b_1 = 1$.
For $k = 2$, notice that $b_2 \le 3$.
Now, suppose that $b_k \le \frac{k(k+1)}{2}$ for a value $k  \ge 2$.
\[  b_{k+1} < \frac{(k+1)^2}{k^2} b_k \le \frac{(k+1)^3}{2k} = \frac{k^2 + 3k + 2}{2} + \frac{k + 1}{2k}\]As $\frac{k^2+3k+2}{2} \in \mathbb{N}$ and $0 < \frac{k+1}{2k} < 1$. This gives us
\[ b_{k+1} \le \left \lfloor \frac{k^2 + 3k + 2}{2} + \frac{k + 1}{2k} \right \rfloor = \frac{(k+1)(k+2)}{2} \]which completes the induction.
To prove that $r = \frac{1}{2}$ is achievable. Take the sequence $b_k = \frac{k(k+1)}{2}$ for all $k \in \mathbb{N}$.
\[ \lim_{n \to \infty} \frac{b_n}{n^2} = \lim_{n \to \infty} \frac{n+1}{2n} = \frac{1}{2} \]
It suffices to prove that for any positive reals $0 < r < \frac{1}{2}$, we can have
\[ \lim_{n \to \infty} \frac{b_n}{n^2} = r \]This is possible by taking $b_n = \lceil rn^2 \rceil + n$ for some $n > N$ and $b_n = \frac{n(n+1)}{2}$, when $n \le N$. We'll first prove that such sequence satisfy.

Now, we'll prove that such sequence $b_k$ satisfies
\[ \frac{b_k}{k^2} > \frac{b_{k+1}}{(k+1)^2} \]By expanding, we want to prove that
\[ (k+1)^2k + (k+1)^2 \lceil rk^2 \rceil > k^2(k+1) + k^2 \lceil r(k+1)^2 \rceil \]\[ k(k+1) + (k+1)^2 \lceil rk^2 \rceil > k^2 \lceil r(k+1)^2 \rceil \]But actually,
\begin{align*}
 (k+1)^2 \lceil rk^2 \rceil &> (k+1)^2 (rk^2) \\ &= k^2 (r(k+1)^2 + 1) - k^2 \\ &> k^2 \lceil r(k+1)^2 \rceil - k^2
\end{align*}which is true.
Now, we need to find a constraint for $N$. Since $b_n = \frac{n(n+1)}{2}$ for all $n \le N$. Then we have $b_{N + 1} < \frac{(N+1)(N+2)}{2}$ as well. This gives us
\[ \lceil r(N+1)^2 \rceil  + N + 1 < \frac{(N+1)(N+2)}{2} \]But for large enough $N$, we must have
\[ \lceil r(N+1)^2 \rceil + N + 1 < r(N+1)^2 + N + 1 < \frac{1}{2} N^2 + \frac{3}{2}N + 1 \]as $r < \frac{1}{2}$.
We are hence finished.
Now,
\[ \lim_{n \to \infty} \frac{b_n}{n^2} = \lim_{n \to \infty} \frac{\lceil rn^2 \rceil + n}{n^2} = r \]because
\[ r = \lim_{n \to \infty} \frac{ rn^2 + n}{n^2} \le \lim_{n \to \infty} \frac{ \lceil rn^2 \rceil + n}{n^2} \le \lim_{n \to \infty} \frac{rn^2 + n + 1}{n^2} = r \]
Motivation
This post has been edited 1 time. Last edited by IndoMathXdZ, Dec 17, 2019, 10:14 AM
Reason: typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
pad
1671 posts
#8 • 3 Y
Y by centslordm, 554183, Adventure10
Solution

Remarks
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
niyu
830 posts
#9 • 2 Y
Y by surferdude11, centslordm
We claim all $0 \leq r \leq \frac{1}{2}$ work.

We first prove that all $r$ are in this range. To do so, we will prove that $b_n \leq \frac{n^2 + n}{2}$ for all $n$. We do so by induction on $n$. As the base case, we have $b_1 = 1 = \frac{1^2 + 1}{2}$. Now, suppose $b_k \leq \frac{k^2 + k}{2}$. We have
\begin{align*}
        \frac{b_{k + 1}}{(k + 1)^2} &< \frac{b_k}{k^2} \\
        &< \frac{k^2 + k}{2k^2} \\
        &< \frac{k + 1}{2k} \\
        b_{k + 1} &< \frac{(k + 1)^3}{2k} \\
        b_{k + 1} &< \frac{k^2}{2} + \frac{3k}{2} + \frac{3}{2} + \frac{1}{2k}.
\end{align*}However, note that
\begin{align*}
        \frac{(k + 1)^2 + (k + 1)}{2} &= \frac{k^2 + 3k + 2}{2} \\
        &< \frac{k^2}{2} + \frac{3k}{2} + \frac{3}{2} + \frac{1}{2k} \\
        \frac{(k + 1)^2 + (k + 1)}{2} + 1 &= \frac{k^2 + 3k + 4}{2} \\
        &\geq \frac{k^2}{2} + \frac{3k}{2} + \frac{3}{2}k + \frac{1}{2k},
\end{align*}which is enough to show that $b_{k + 1} \leq \frac{(k + 1)^2 + (k + 1)}{2}$, completing the induction.

Hence, we have $\frac{b_n}{n^2} \leq \frac{1}{2} + \frac{1}{2n}$. As $n$ approaches infinity, the right side approaches $\frac{1}{2}$, showing that $r \leq \frac{1}{2}$. Clearly $\frac{b_n}{n^2} \geq \frac{1}{n^2}$, which approaches $0$ as $n$ approaches infinity. Hence, $r \geq 0$, showing that $0 \leq r \leq \frac{1}{2}$.

We now show that all $0 < r < \frac{1}{2}$ work (we have already provided constructions for $r = 0, \frac{1}{2}$). Consider some fixed $r$, and the sequence $b_n$ for which $b_n = \frac{n^2 + n}{2}$ if $rn^2 + n < \frac{n^2 + n}{2} + 100$ (this is false for large enough $n$ since $r < \frac{1}{2}$), and $b_n = \lceil rn^2 + n \rceil$ otherwise. This sequence satisfies $b_n \leq \frac{n^2 + n}{2}$ for all $n$ (which is necessary as $\frac{n^2 + n}{2}$ is the maximum value of $b_n$). We now show that
\begin{align*}
        \frac{b_n}{n^2} &> \frac{b_{n + 1}}{(n + 1)^2}
\end{align*}for all $n$. If $b_n = \frac{n^2 + n}{2}$ and $b_{n + 1} = \frac{(n + 1)^2 + (n + 1)}{2}$ this is true (as checked above). Otherwise, if $b_n = \frac{n^2 + n} {2}$ and $b_{n + 1} = \lceil r(n + 1)^2 + (n + 1) \rceil$, this is true because we have
\begin{align*}
        \frac{b_n}{n^2} &= \frac{\frac{n^2 + n}{2}}{n^2} \\
        &> \frac{\frac{(n + 1)^2 + (n + 1)}{2}}{(n + 1)^2} \\
        &> \frac{\lceil r(n + 1)^2 + (n + 1) \rceil}{(n + 1)^2} \\
        &= \frac{b_{n + 1}}{(n + 1)^2}.
\end{align*}Finally, suppose $b_n = \lceil rn^2 + n \rceil$ and $\lceil r(n + 1)^2 + (n + 1) \rceil$. We have
\begin{align*}
        \frac{rn^2 + n}{n^2} &> \frac{r(n + 1)^2 + (n + 1) + 1}{(n + 1)^2} \\
        \iff \frac{1}{n} &> \frac{n + 2}{(n + 1)^2} \\
        \iff (n + 1)^2 &> n(n + 2),
\end{align*}which is true. Thus, we have
\begin{align*}
        \frac{\lceil rn^2 + n \rceil}{n^2} &\geq \frac{rn^2 + n}{n^2} \\
        &> \frac{r(n + 1)^2 + (n + 1) + 1}{(n + 1)^2} \\
        &> \frac{\lceil r(n + 1)^2 + (n + 1) \rceil}{(n + 1)^2},
\end{align*}or $\frac{b_n}{n^2} > \frac{b_{n + 1}}{(n + 1)^2}$. Thus, this sequence satisfies the given condition. As the infimum of $\frac{b_n}{n^2}$ for this sequence is $r$, we may conclude that all $0 \leq r \leq \frac{1}{2}$ are achievable, as claimed.

This completes the proof.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
stamatelos
17 posts
#10 • 2 Y
Y by centslordm, Mango247
What is the logic behind bn<n(n+1)/2? you find it by try and error or its a method yo find it?
This post has been edited 2 times. Last edited by stamatelos, Jun 29, 2020, 8:00 PM
Reason: typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
jj_ca888
2726 posts
#11 • 3 Y
Y by centslordm, Mango247, Mango247
stamatelos wrote:
What is the logic behind bn<n(n+1)/2? you find it by try and error or its a method yo find it?

Short answer: Trial and Error

Long answer: I think this is quite intuitive. When I did this problem I first listed out values of $b_i$'s.

Note that $b_1 = 1$. Then, we need $\tfrac{b_2}{4} < 1$ so the maximum possible value of $b_2$ is $3$. Then, we need $\tfrac{b_3}{9} < \tfrac34$ which yields the maximum possible value of $b_3$ is $6$. Then, we need $\tfrac{b_4}{16} < \tfrac69$ so the maximum possible value (after some computation) of $b_4$ is $10$.

So the maximum possible value of the first four $b_i$'s follows the sequence $1, 3, 6, 10$. Hopefully this looks familiar. After noting the (quite obvious) pattern at this point, you should be ready to induct.
This post has been edited 4 times. Last edited by jj_ca888, Jun 29, 2020, 8:06 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
smartninja2000
1631 posts
#13 • 1 Y
Y by centslordm
Wait, this seems similar to some CMC 10A problem...
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
arvind_r
136 posts
#14 • 1 Y
Y by centslordm
Why does $r < 0$ not work? (i.e. what if the $b_i$'s are negative?)
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
GorgonMathDota
1063 posts
#15 • 3 Y
Y by cosmicgenius, arvind_r, centslordm
arvind_r wrote:
Why does $r < 0$ not work? (i.e. what if the $b_i$'s are negative?)


a1267ab wrote:
Choose positive integers $b_1, b_2 \dots$
Your welcome
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathlogician
1051 posts
#16 • 2 Y
Y by centslordm, Mango247
The answer is $0 \leq r \leq 1/2$.

Proof of Necessity: Choose the $(b_i)$ to be as large as possible. Now I claim that $b_n = n(n+1)/2$ for all positive integers $n$, by induction. Note that if $b_n = \frac{n(n+1)}{2}$, it remains to show that $b_{n+1} = \frac{(n+1)(n+2)}{2}$ works but $b_{n+1} = \frac{(n+1)(n+2)}{2}+1$ fails. Therefore, it suffices to show $$\frac{(n+1)(n+2)}{(n+1)^2} \leq\frac{n(n+1)}{n^2} \leq \frac{(n+1)(n+2)+2}{(n+1)^2}.$$
The left inequality expands to $n(n+2)\leq (n+1)^2$, while the right inequality expands to $(n+1)^3 \leq n(n^2+3n+4)$, or $n^3+3n^2+3n+1 \leq n^3+3n^2+4n \implies 1 \leq n$, obvious.

Now obviously $\frac{b_n}{n^2} \leq \frac{n(n+1)}{2n^2} = \frac{n+1}{2n}$, so $r \leq 1/2$.

Construction: It remains to show that any $r$ for $0 \leq r \leq 1/2$ is achievable for some choice of $(b_i)$. Set $b_n = \left\lceil rn^2 + n\right\rceil$ if $\frac{\left\lceil rn^2 + n\right\rceil}{n^2} < \frac{1}{2}$, and set $b_n = n(n+1)/2$ otherwise. It suffices to show that $b_n/n^2 > b_{n+1}/(n+1)^2$, as this sequence will tend towards $r$ for large $n$. One may manually check that this construction works, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
IAmTheHazard
5005 posts
#17 • 1 Y
Y by centslordm
The answer is $r \in [0,\tfrac{1}{2}]$.
I will first prove necessity. Clearly, $r \geq 0$, so we only have to prove $r \leq \tfrac{1}{2}$. This is established with the following claim.

Claim: For all $n$, we have $b_n\leq \tfrac{1}{2}n(n+1)$.
Proof: Use induction on $n$, with the base case of $n=1$ being clear. Suppose now that we have $b_n \leq \tfrac{1}{2}n(n+1)$. I will show that $b_{n+1} \leq \tfrac{1}{2}n(n+1)$. We require:
$$\frac{b_n}{n^2}>\frac{b_{n+1}}{(n+1)^2} \implies \frac{\tfrac{1}{2}n(n+1)}{n^2}=\frac{n+1}{2n}>\frac{b_{n+1}}{(n+1)^2}.$$From here, it's not hard to verify that all $b_{n+1} \geq \tfrac{1}{2}n(n+1)+1$ fail this requirement, thus completing the induction. This clearly implies $r\leq \tfrac{1}{2}$.

It remains to provide a construction. For $r=\tfrac{1}{2}$, we can take $b_n=\tfrac{1}{2}n(n+1)$, which gives $\tfrac{b_n}{n}=\tfrac{1}{2}+\tfrac{1}{2n}$ for all $n$. This is clearly valid.
Now we deal with $r<\tfrac{1}{2}$. For some arbitrary $r \in [0,\tfrac{1}{2})$, consider the sequence $(a_n)$ defined by $a_n=\lceil rn^2\rceil+n$. It is clear that for sufficiently large $n$, we have
$$a_n\leq rn^2+n+1<\tfrac{1}{2}n(n+1).$$So we can take some positive integer $N$ such that for all $N\geq n$, $a_n<\tfrac{1}{2}n(n+1)$. Then define
$$b_n=\begin{cases} \frac{1}{2}n(n+1)& n<N\\ a_n & n\geq N.\end{cases}$$Observe that
$$\frac{b_n}{n^2}\geq \frac{a_n}{n^2}=\frac{\lceil rn^2\rceil+n}{n^2}\geq \frac{rn^2}{n^2}=r,$$so we have $\tfrac{b_n}{n^2} \geq r$ for all $n \geq 1$. Since $\lim_{n \to \infty} \tfrac{b_n}{n^2}=r$, it follows that $r$ is maximal. Hence we only have to verify that $\tfrac{b_n}{n^2}>\tfrac{b_{n+1}}{(n+1)^2}$ for all $n \geq 1$. This was already proven for all $n<N-1$ and is clear for $n=N-1$, so we only have to prove it for $n \geq N$. It suffices to show that
$$\frac{a_n}{n^2}>\frac{a_{n+1}}{(n+1)^2} \iff \frac{\lceil rn^2\rceil+n}{n^2}>\frac{\lfloor r(n+1)^2\rfloor+n}{(n+1)^2}$$holds for all $n \geq N$. We have:
\begin{align*}
\frac{\lceil rn^2\rceil+n}{n^2}&>\frac{\lceil r(n+1)^2\rceil+(n+1)}{(n+1)^2}&&\iff\\
(n+1)^2\lceil rn^2\rceil+n(n+1)^2&>n^2\lceil r(n+1)^2\rceil +n^2(n+1)&&\iff\\
n^2+n&>n^2\lceil r(n+1)^2\rceil-(n+1)^2\lceil rn^2\rceil&&\iff\\
n^2+n&>n^2(r(n+1)^2+\{1-r(n+1)^2\})-(n+1)^2(rn^2+\{1-rn^2\})&&\iff\\
n^2+n&>n^2\{1-r(n+1)^2\}-(n+1)^2\{1-rn^2\},
\end{align*}where we use the easily verifiable identity $\lceil x \rceil=x+\{1-x\}$ to get from the third line to the fourth.
Note that we have $n^2\{1-r(n+1)\}<n^2$, and $(n+1)^2\{1-rn^2\}$ must be nonnegative, so
$$n^2\{1-r(n+1)^2\}-(n+1)^2\{1-rn^2\}>n^2.$$As $n^2+n>n^2$, the original inequality is true, so we indeed have $\tfrac{b_n}{n^2}>\tfrac{b_{n+1}}{(n+1)^2}$ for all $n$. Hence, this construction for $r$ works, and we're done. $\blacksquare$
This post has been edited 1 time. Last edited by IAmTheHazard, Jun 7, 2021, 6:45 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
somewhere123
2 posts
#18
Y by
为什么我无法下载这个文档,帮帮我
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Apple321
1506 posts
#19
Y by
somewhere123 wrote:
为什么我无法下载这个文档,帮帮我

Why can't you download the document?

I'm not sure what document your talking about..
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
508669
1040 posts
#20
Y by
a1267ab wrote:
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?

Carl Schildkraut and Milan Haiman

Posting for storage.

We see that if $b_n \leq \frac{n(n+1)}{2}$, then $b_{n+1}^2 < \dfrac{(n+1)^2b_n}{n^2} \leq \frac{(n+1)^3}{2n} = \dfrac{n^3 + 3n^2 + 3n + 1}{2n} = \dfrac{(n+1)(n+2)}{2} + \frac{n+1}{2n} \implies b_{n+1} \leq \dfrac{(n+1)(n+2)}{2}$ and $b_1 = \frac{1 \cdot 2}{2}$. This means that $\frac{b_n}{n^2} \leq \frac{n+1}{2n}$ which can be $\frac{1}{2} + \epsilon$ where $\epsilon$ is an arbitrarily small positive real number. This means that $r \leq \frac{1}{2}$.

We claim that all reals $r \in [0, \frac{1}{2}]$ work. Simply choose $b_n = \lceil rk^2 + k \rceil$ if $\lceil rk^2 + k \rceil < \frac{k^2}{2}$ or $b_k = \frac{k(k+1)}{2}$ otherwise. Simply because $\lceil rk^2 + k \rceil < \frac{k^2}{2}$ may not be true for all positive integers $k$. We can see that this construction indeed works.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
FishHeadTail
75 posts
#21
Y by
I wonder what’s the motivation behind looking at $\lceil cn^2 \rceil +n$. Thanks a lot!
This post has been edited 3 times. Last edited by FishHeadTail, Sep 2, 2021, 2:07 PM
Reason: Typo
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
mathlogician
1051 posts
#22
Y by
FishHeadTail wrote:
I wonder what’s the motivation behind looking at $\lceil cn^2 \rceil +n$. Thanks a lot!

Here's how I came up with it (might be already mentioned in the thread). We need $cn^2$ so the limit of $(b_n)$ approaches $c$. However $c$ can be any real number, so we use the ceiling. However, $b_n = \lceil{cn^2 \rceil}$ still doesn't work, so we can add a linear term, knowing that the limit is still $c$ but $\tfrac{b_n}{n^2}$ is decreasing. There is one more issue: the initial terms are too large, but this is an easy fix as we can just replace them with triangular numbers, the end.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
guillermo.dinamarca
1 post
#23
Y by
a1267ab wrote:
Choose positive integers $b_1, b_2, \dotsc$ satisfying
\[1=\frac{b_1}{1^2} > \frac{b_2}{2^2} > \frac{b_3}{3^2} > \frac{b_4}{4^2} > \dotsb\]and let $r$ denote the largest real number satisfying $\tfrac{b_n}{n^2} \geq r$ for all positive integers $n$. What are the possible values of $r$ across all possible choices of the sequence $(b_n)$?

Carl Schildkraut and Milan Haiman
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
megarnie
5611 posts
#24
Y by
Note $b_1=1$.

We claim the answer is $\boxed{0\le r\le \frac{1}{2}}$.

Part 1: Show that $r=\frac{1}{2}$ works, and that it is maximal.
For $\frac{1}{2}$, set $b_n=\frac{n^2+n}{2}$, which is always an integer. So $\frac{b_n}{n^2}=\frac{1}{2}+\frac{1}{2n}$. The sequence $\frac{1}{2n}$ will converge to $0$, so $\frac{b_n}{n^2}$ will converge to $\frac{1}{2}$.

Now we will show that $\frac{1}{2}$ is maximal. We will use the following claim.

Claim: $\frac{b_n}{n^2}\le \frac{1}{2}+\frac{1}{2n}$, which obviously proves the first part.
Proof: We will use induction.
Base case(s): $n=1,2$ ($n=2$ because the maximal value for $\frac{b_2}{2^2}$ is $\frac{3}{4}=\frac{1}{2}+\frac{1}{4})$.

Inductive step: Suppose $\frac{b_k}{k^2}\le \frac{1}{2}+\frac{1}{2k}\forall k<n$. Then we suppose for the sake of contradiction that $\frac{b_n}{n^2}>\frac{1}{2}+\frac{1}{2n}$. Since $b_n$ and $n$ are both positive integers, the minimum value for $\frac{b_n}{n^2}$ is $\frac{1}{2}+\frac{1}{2n}+\frac{1}{n^2}$.

This gives us the inequality $\frac{1}{2}+\frac{1}{2n}+\frac{1}{n^2}<\frac{1}{2}+\frac{1}{2n-2}\implies \frac{1}{2n}+\frac{1}{n^2}=\frac{n+2}{2n^2}<\frac{1}{2n-2}$. Multiplying both sides by $2n^2$ gives $n+2<\frac{n^2}{n-1}$. Since $n>1$, multiplying both sides by $n-1$ gives $(n+2)(n-1)<n^2\implies n^2+n-2<n^2\implies n-2<0$, a contradiction as $n\ge 2$.



Part 2: Show that all $0\le r<\frac{1}{2}$ work.
Obviously we can set $b_i=i\forall i$, which gives $r=0$, so henceforth assume $0<r<\frac{1}{2}$.

Let $N$ be a sufficiently large value so that for all $k\ge N$, $\left\lceil rk^2+k\right\rceil<\frac{k^2+k}{2}$.

Let $b_k=\left\lceil rk^2+k \right\rceil\forall k\ge N$ and $b_n=\frac{n^2+n}{2}\forall k<N$.

Clearly $\frac{b_k}{k^2}$ converges to $\frac{rk^2+k+c}{k^2}=r+\frac{1}{k}+\frac{c}{k^2}$, where $c<1$. Since both $\frac{1}{k}$ and $\frac{c}{k^2}$ converge to $0$, $\frac{b_k}{k^2}$ converges to $r$. It suffices to show that it's strictly decreasing.

For all $n<N$, $\frac{b_n}{n^2}=\frac{1}{2}+\frac{1}{2n}$, which is strictly decreasing.

We note $\frac{b_{N-1}}{(N-1)^2}=\frac{1}{2}+\frac{1}{2N-2}>\frac{1}{2}+\frac{1}{2N}>\frac{b_{N}}{N^2}$.

Thus, it suffices to show that for all $k\ge N$, $\frac{b_k}{k^2}>\frac{b_{k+1}}{(k+1)^2}$. We have \[\frac{b_k}{k^2}=\frac{\left\lceil rk^2+k\right\rceil}{k^2}\ge\frac{rk^2+k}{k^2}\]
Claim: $\frac{rk^2+k}{k^2}>\frac{r(k+1)^2+(k+1)+1}{(k+1)^2}$. If we have proven this, we are done as $\frac{r(k+1)^2+(k+1)+1}{(k+1)^2}>\frac{b_{k+1}}{(k+1)^2}$
Proof: AFTSOC $\frac{rk^2+k}{k^2}\le \frac{r(k+1)^2+(k+1)+1}{(k+1)^2}$. Then $r+\frac{1}{k}\le r+\frac{1}{k+1}+\frac{1}{(k+1)^2}\implies \frac{1}{k}\le \frac{1}{k+1}+\frac{1}{(k+1)^2}=\frac{k+2}{(k+1)^2}$.

Multiplying both sides by $k(k+1)^2$ gives $(k+1)^2\le k(k+2)\implies k^2+2k+1\le k^2+2k$, which is absurd.
This post has been edited 1 time. Last edited by megarnie, Dec 28, 2021, 8:23 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
asdf334
7585 posts
#25
Y by
After getting the upper and lower bounds on $r$, another approach would be to pick any arbitrary value of $r$ and an arbitrarily large $k$, then set $b_k$ as the smallest integer satisfying $\frac{b_k}{k^2}>r$ and move "backward", picking the smallest fraction greater than the previous, and show that this works via contradiction.
This post has been edited 2 times. Last edited by asdf334, Jan 15, 2022, 3:44 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Inconsistent
1455 posts
#26
Y by
The answer is all reals between $0$ and $\frac{1}{2}$. Upper bound is trivial by thinking. Construction is to stay on the upper bound construction until you are able to switch to $b_n = \lceil rn^2 \rceil + n$, finishing.
This post has been edited 1 time. Last edited by Inconsistent, Oct 27, 2022, 10:03 PM
Reason: edit
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
EthanWYX2009
872 posts
#27
Y by
引理: 对于 $\forall n\in\mathbb Z_+$, 都有 $b_n\leqslant\frac 12n(n+1)$.
我们运用数学归纳法证明该引理. 由已知 $n=1$ 时结论成立. 假设对于 $n>1$, 结论对于 $n-1$ 成立, 则有
$$b_n<\frac{n^2}{(n-1)^2}b_{n-1}\leqslant\frac{n^2}{(n-1)^2}\cdot\frac 12n(n-1)=\frac{n^3}{2(n-1)}<\frac{n(n+1)}{2}+1$$结合 $b_n\in\mathbb Z$,$b_n\leqslant\frac 12n(n+1)$, 归纳成立.
回到原题, 由引理知 $r\leqslant\frac{b_n}{n^2}\leqslant\frac 12+\frac 1{2n}$, 因此 $r\leqslant\frac 12$. 对于数列 $b_n=\frac 12n(n+1)$,$r=\frac 12$; 对于数列 $b_n\equiv 1$,$r=0$.
对于 $0<r<\frac 12$,$N\in\mathbb Z_+$, 使得 $n\geq N$ 时, 都有 $\left\lceil rn^2+n\right\rceil <\frac 12n(n+1)$.
取数列 $b_n=\frac 12n(n+1)$, $1\leq n<N$; $b_n=\left\lceil rn^2+n\right\rceil$, $n\geq N$.$\lim_{n\to +\infty}\frac {b_n}{n^2}=r$.
综上所述, ${r}$ 的取值范围为 $\boxed{\left[0,\frac 12\right]}$.$\blacksquare$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
john0512
4191 posts
#29
Y by
We claim that the answer is $0\leq r\leq 1/2.$

Claim 1: $$b_n\leq \frac{n(n+1)}{2}.$$We will use induction. Clearly, this is true for $n=1,2,3.$ We will use induction. Suppose that $$b_k\leq \frac{k(k+1)}{2}$$for some $k\geq 3$. Then, $$b_{k+1}<b_k\frac{(k+1)^2}{k^2}\leq \frac{(k+1)^3}{2k}.$$
Case 1: $k$ is even. Then, $$\frac{(k+1)^3}{2k}=\frac{k^2}{2}+\frac{3k}{2}+\frac{3}{2}+\frac{1}{2k}.$$The first two terms will be integers if $k\geq 3$ and $k$ is even, so its floor is $$\frac{k^2}{2}+\frac{3k}{2}+1=\frac{(k+1)(k+2)}{2},$$and since $$b_{k+1}\leq \lfloor \frac{(k+1)^3}{2k}\rfloor,$$this case is resolved.

Case 2: $k$ is odd. Then, let $k=2s-1$, so $$b_{k+1}\leq \lfloor \frac{(k+1)^3}{2k}\rfloor =\lfloor \frac{4s^3}{2s-1}\rfloor=\lfloor 2s^2+s+\frac{1}{2}+\frac{1}{4s-2}\rfloor=2s^2+s,$$which is what we want. Hence, we have shown the claim.

This clearly shows that $r\leq 1/2$. It also shows that $r=1/2$ is achievable since we can just set $b_n=\frac{n(n+1)}{2}.$

Clearly, $r\geq 0$. Furthermore, $r=0$ achievable by $b_n=1$. It remains to show that $0<r<1/2$ is achievable.

Claim 2: If $0<r<1/2$ is a real number, then $$\frac{\lceil rn^2 \rceil +n}{n^2}$$is decreasing with respect to $n$ when $n$ is a positive integer. This is just showing that $$\frac{\lceil rn^2 \rceil +n}{n^2}>\frac{\lceil r(n+1)^2 \rceil +n+1}{(n+1)^2}.$$Note that we have $$\lceil rn^2 \rceil\geq rn^2$$and $$\lceil r(n+1)^2\rceil < r(n+1)^2+1,$$so it suffices to show that $$\frac{rn^2 +n}{n^2}>\frac{r(n+1)^2+1 +n+1}{(n+1)^2}.$$This is just $$\frac{1}{n}>\frac{n+2}{(n+1)^2}$$$$(n+1)^2>n(n+2),$$which is clearly true.

Note that $$\lim_{n\rightarrow \infty}\frac{\lceil rn^2 \rceil +n}{n^2}=r,$$and furthermore, $$\frac{\lceil rn^2 \rceil +n}{n^2}>r$$for all $n$. Thus, if $r<1/2$, we can first do $b_n= \frac{n(n+1)}{2}$ for sufficiently many terms, and then swap over to $$b_n=\frac{\lceil rn^2 \rceil +n}{n^2}$$and do that for the rest of the way to achieve $r$ (this works if we go sufficiently far since it goes from larger to 1/2 to below 1/2 during the transition if we wait sufficiently long, since it heads towards $r<1/2$), so we are done.
This post has been edited 1 time. Last edited by john0512, May 21, 2023, 9:30 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
naonaoaz
334 posts
#31
Y by
Obviously $r \ge 0$, with equality achievable by $b_i = 1$ for all $i$.

Similarly, note that $r \le \frac{1}{2}$. To see this, we can imagine starting with $b_1 = 1$ and greedily picking the largest possible $b_2,b_3,\ldots$. It's clear this greedy strategy will give the largest possible $r$.

Using this greedy method, induction shows that $b_n = {{n+1} \choose 2}$ for all $n \ge 1$. Then taking $\lim_{n \to \infty} \frac{b_n}{n^2} = \frac{1}{2}$ finishes. Furthermore, this implies $b_n \le {{n+1} \choose 2}$ for any sequence $b_i$.
Claim: All $r \in \left[0,\frac{1}{2}\right]$ are achievable.
Proof: We've already shown $0$ and $\frac{1}{2}$ are achievable. Consider $a_n = \left \lceil{rn^2}\right \rceil+n$. Clearly, $\lim_{n \to \infty} \frac{a_n}{n^2} = r$. We claim that
\begin{align*}
        b_n &= {{n+1} \choose 2} \text{ for $n \le N$} \\
        b_n &= \left \lceil{rn^2}\right \rceil+n \text{ else}
    \end{align*}for some sufficiently large $N$ works as a sequence. First, to determine $N$, just take any $n$ such that
\[\frac{1}{1-2r} < \frac{n^2}{n+2} \text{ which implies } \left \lceil{rn^2}\right \rceil+n < \text{max $b_n$} = {{n+1} \choose 2}\]Secondly, it's not hard to verify that, when $n>N$, the sequence $\frac{b_n}{n^2}$ is decreasing as desired. Thus since these two conditions are met, this sequence $b_n$ works, and we're done. $\square$
Remark:
The actually checking of the inequality is omitted as it's not difficult or useful. However, a small note: to actually verify the inequalities, use $\left \lceil{x}\right \rceil \ge x$ and $ x+1 \ge \left \lceil{x}\right \rceil$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
YaoAOPS
1541 posts
#32
Y by
Note that $b_i = 1$.
We have that for $j > i$ \[ b_{j} \le \left\lfloor \frac{(j)^2}{i^2} \cdot b_j \right\rfloor \]Note that the inequality is the tightest when $j = i + 1$
Consider the maximal possible value of $r$, which occurs when equality holds between $i, j = i + 1$. We claim that this value is $\frac{1}{2}$.
We have that \[ b_{i+1} = \left\lfloor \frac{(i+1)^2}{i^2} \cdot b_i \right\rfloor = b_i + \left\lfloor \frac{2}{i} b_i \right\rfloor \]Thus, if $i \mid 2b_i$, then $b_{i+1} = \frac{i + 2}{i}b_i$.
Since $b_1 = 1$, we can inductively solve to get $b_2 = 3$, $b_3 = 6$, $b_n = \frac{n(n+1)}{2}$ and as $n \to \infty$, $\frac{b_n}{n^2} \to \frac{1}{2}$.
Claim: If $r$ is the maximal for a fixed $b_i$, then $\frac{b_i}{i^2} - r \le C_j = \sum_{j=i}^{\infty} \frac{1}{j^2}$
Proof. Take $b_{i+1}$ as maximal, repeat to get a decrease of at most $C_j$. $\blacksquare$
Now, define $b_i$ inductively as maximal values such that $\frac{b_i}{i^2} < \frac{b_{i-1}}{(i-1)^2}$, jumping down to $b_i = \left\lceil i^2 (r + C_i) \right\rceil$ whenever $\frac{b_{i-1}}{(i-1)^2} > \frac{1}{i-1} + r + C_i$.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
cursed_tangent1434
649 posts
#33 • 1 Y
Y by GeoKing
We claim that the only real constants $r$ for which such a sequence of positive integers exist are $0 \le r \le \frac{1}{2}$. We start off with proving the bound.

It is not hard to see that $r \ge 0$ since all the terms of the form $\frac{b_i}{i^2}$ are strictly positive. For the upper bound, we first note that, $b_1=1$. Further, we can show the following result via induction.

Claim : For all positive integers $i \ge 2$,
\[\frac{b_i}{i^2}\le \frac{i+1}{2i}\]

Since $b_1=1$ and $\frac{b_2}{4} < 1$ we have $b_2 <4$ and thus, $b_2 \le 3$ implying $\frac{b_2}{4} \le \frac{3}{4}$ as desired. Now, we assume that for some positive integer $k \ge 2$, $\frac{b_k}{k^2} \le \frac{k+1}{2k}$. Then,
\begin{align*}
\frac{b_{k+1}}{(k+1)^2} & < \frac{b_k}{k^2} \\
b_{k+1} & < \frac{(k+1)^3}{2k}\\
b_{k+1} & \ge \frac{(k+1)^3-1}{2k}\\
&= \frac{(k+1)^2+(k+1)+1}{2}\\
b_{k+1} & \ge \frac{k^2+3k+2}{2}\\
&= \frac{(k+1)(k+2)}{2}
\end{align*}using the fact that $b_{k+1} \in \mathbb{N}$. Thus, $b_{k+1} \le \frac{(k+1)(k+2)}{2}$ from which it follows that, $\frac{b_{k+1}}{(k+1)^2} \le \frac{k+2}{2(k+1)}$ completing the induction.

Now, if $r>\frac{1}{2}$, there must exist some $\epsilon >0$ and $k \in \mathbb{N}$ such that for all $i \ge k$, $b_i \ge \frac{1}{2} + \epsilon$. But now, considering $i > \frac{1}{2\epsilon}$ contradicts the above claim, which finishes the proof of the bound.

All that remains now is to provide a construction. When $r=0$ and $r=\frac{1}{2}$ simply consider the sequences $b_i=i$ and $b_i = \frac{i(i+1)}{2}$ respectively. For all $0 < r < \frac{1}{2}$ we can consider the sequence,
\[b_i= \begin{cases}
\frac{i(i+1)}{2} & i < N\\
\lceil ri^2+i \rceil & i \ge N
\end{cases}\]for sufficiently large $N$. To see why this works we let $c_i = \frac{b_i}{i^2}$ for all positive integers $i$, it is first clear that $c_1=1$ and $c_i$ is increasing for $1 \le i <  \frac{3}{1-2r}$. Then, we have two consecutive terms of the form,
\[c_{k-1} = \frac{k}{2(k-1)} \text{ and }  c_k = \frac{\lceil rk^2+k \rceil}{k^2}\]Note that,
\begin{align*}
c_k & = \frac{\lceil rk^2+k \rceil}{k^2} \\
& \le \frac{rk^2+k+1}{k^2}\\
& < \frac{k}{2(k-1)}\\
&= c_{k-1}
\end{align*}for sufficiently large $k$ (so we simply need to select $N$ such that the final inequality holds). Further, for all $i>k$, $c_i$ is also increasing since,
\begin{align*}
\frac{\lceil ri^2+i \rceil}{i^2} & >  \frac{ri^2 +i}{i^2}\\
& = r + \frac{1}{i}\\
& > r + \frac{i+2}{(i+1)^2}\\
& > \frac{r(i+1)^2 + (i+1) + 1}{(i+1)^2}\\
& > \frac{\lceil r(i+1)^2+ (i+1) \rceil}{(i+1)^2}
\end{align*}Thus, the described sequence satisfies all the desired characteristics. Further,
\[\frac{b_n}{n^2} = \frac{\lceil rn^2+n \rceil}{n^2}> r\]So, $r$ is a lower bound of $c_i$. To see why it is the greatest lower bound, say there exists some $\epsilon >0$ and $k \in \mathbb{N}$ such that for all $ i \ge k$ we have $c_i \ge r+ \epsilon$. Then, we have $\frac{\lceil ri^2+i \rceil}{i^2} > r + \epsilon$ so,
\begin{align*}
ri^2+i+1 & > \lceil ri^2+i \rceil > ri^2 + i^2 \epsilon\\
i+1 & > i^2 \epsilon
\end{align*}which is clearly false for sufficiently large $i$. Thus, $r$ is in fact the greatest lower bound of $c_i$ which completes the solution.
This post has been edited 1 time. Last edited by cursed_tangent1434, Jul 4, 2024, 5:48 AM
Reason: typoes
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Ywgh1
139 posts
#34
Y by
USA 2020 TST p1

We claim that $r \in [0,1/2]$.
We start off with the following claim.

Claim : $b_n \leq \frac{n(n+1)}{2}$ for all $n$.

Proof: We use induction, base case being $n=1$ is trivial. First assume that $b_{n-1} \leq \frac{n(n-1)}{2}$,
we show that $b_n \leq \frac{n(n+1)}{2}$.

\begin{align*}
b_{n} &<\frac{n(n-1)}{2}\cdot\frac{n^2}{(n-1)^2} \\
&= \frac{n^3-n}{2(n-1)} + \frac{n}{2(n-1)} \\
&< \frac{n(n+1)}{2}+1
\end{align*}As desired. $\blacksquare$

Now as $n \to \infty$ we get that
\[\frac{b_n}{n^2} \leq \frac{1}{2}\].

Now we give a construction of our bound.

Construction: Let $b_n = \left\lceil rn^2 + n\right\rceil$ if $\frac{\left\lceil rn^2 + n\right\rceil}{n^2} < \frac{1}{2}$, otherwise, let $b_n=\frac{n(n+1)}{2}$ .
Which works, hence we are done.
This post has been edited 5 times. Last edited by Ywgh1, Aug 15, 2024, 7:30 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
ihatemath123
3449 posts
#35
Y by
The answer is $r \in [0, \tfrac{1}{2} ]$.

Claim: We have $b_i \leq \tfrac{i(i+1)}{2}$.
Proof: We prove this with induction, with the base case of $i=1$ being obvious. For $i > 1$, we have
\[b_i < \frac{i^2}{(i-1)^2} \cdot b_{i-1} \leq \frac{i^3}{2(i-1)} < \frac{i^2 + i + 2}{2},\]and since $\tfrac{i^2 + i + 1}{2}$ is not an integer, it follows that $b_i \leq \tfrac{i(i+1)}{2}$, as claimed.

So, we must have $r \leq \tfrac{1}{2}$. We now show that we can obtain every $r$ in this range. Note that the value of $\tfrac{b_i}{i^2}$ decreases by at most $\tfrac{1}{i^2}$ each time we increment $i$ by one. Therefore, if we define \[f(n) := \sum_{j=n}^\infty \frac{1}{j^2},\]we can always make our sequence converge to some real number at least $L_n = \tfrac{b_n}{n^2} - f(n+1)$. Now, we construct our sequence $b_i$ as follows: for each $i$, first, set each $b_i$ to be as large as possible until $L_i$ is greater than $r$ – this must eventually happen since $\lim_{i \to \infty} f(i) = 0$. Let the $i$-value at which this happens be $k$. We continue to increase $i$, making $b_i$ as large as possible – as we do so, the value of $L_i$ increases. We repeat this until $\tfrac{1}{i^2} < (L_k - r)/2$. (Note the $k$ subscript.) Next, instead of picking $b_i$ to be as large as possible, we first set it to its maximum value and then decrease it by $1$ until $L_i$ lies in the range $(r, (r+L_k)/2)$. (This is possible since, by assumption, $\tfrac{1}{i^2} < (L_k - r)/2$.) Now, we reset $k$ to be the current value of $i$ and repeat this process indefinitely. By doing this, $L_i - r$ approaches $0$, and since $\tfrac{b_i}{i^2} - L_i$ also approaches $0$, it follows that $\tfrac{b_i}{i^2}$ approaches $r$, as desired.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Mathandski
773 posts
#36
Y by
Subjective Rating (MOHs) $       $
Attachments:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Trasher_Cheeser12321
14 posts
#37
Y by
In order to find the maximum possible value for $r$, we need to try maximizing each value in the sequence.

Claim. $b_n$ must be the $n^\text{th}$ triangular number in order for the fraction to be maximized.

Proof. This can be proven using induction with our base cases being $b_1 = 1$ and $b_2 = 1+2 = 3$. With our inductive hypothesis, assume that $b_n = \frac{n(n+1)}{2}$. Using the condition given in the problem, $b_{n+1}$ must satisfy
\[ \frac{\frac{n(n+1)}{2}}{n^2} > \frac{b_{n+1}}{(n+1)^2} \]It can be easily verified that the inequality holds for $b_{n+1} = \frac{(n+1)(n+2)}{2}$. Now all there is left to show is that the condition doesn't hold up when $b_{n+1} = \frac{(n+1)(n+2)}{2}+1$. This can be shown with
\begin{align*}
\frac{n+1}{2n} &> \frac{\frac{(n+1)(n+2)}{2}+1}{(n+1)^2}\\
\frac{n+1}{2n} &> \frac{n^2+3n+2+2}{2(n+1)^2}\\
(n+1)^3 &> n^3 + 3n^2 + 4n
\end{align*}which is false since the statement simplifies to $1>n$ which is absurd. $\blacksquare$

Since $r$ is maximized when $b_n = \frac{n(n+1)}{2}$, we have that
\[ r \le \frac{n+1}{2n} \]for all $n$. As $n$ approaches infinity, we can conclude that $r \le \frac{1}{2}$. Since obviously $r\ge 0$, $r$ must lie in the interval $\left[0, \frac{1}{2}\right]$. The construction for any $r$ in this interval is done by defining
\[ b_n = \begin{cases} \frac{n(n+1)}{2} & \text{if } n \le N \\ \bigl{\lceil}rn^2 + n\bigl{\rceil} & \text{if } n > N \end{cases} \]for a sufficiently large $N$ satisfying $\frac{N(N+1)}{2} > \bigl{\lceil}rN^2 + N\bigl{\rceil} $. Lastly, since
\begin{align*}
\frac{\left\lceil rn^2 + n \right\rceil}{n^2} &\ge r + \frac{1}{n} > r + \frac{n+2}{(n+1)^2} = \frac{\left(r(n+1)^2 + (n+1)\right) + 1}{(n+1)^2} > \frac{\left\lceil r(n+1)^2 + (n+1) \right\rceil}{(n+1)^2}
\end{align*}we see that the condition $b_{n+1}>b_n$ still holds even for $n > N$ and the sequence approaches $r$ as $n$ tends to infinity. $\blacksquare$
This post has been edited 2 times. Last edited by Trasher_Cheeser12321, Dec 26, 2024, 5:49 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aliz
163 posts
#38
Y by
The answer is $\boxed{0 \le r \le \frac{1}{2}}$. Since $\frac{b_n}{n^2} \ge 0$, $r \ge 0$.

Claim: $b_n \le \frac{n(n+1)}{2}$
Proof: We will prove by induction. This is obvious for $n = 1$. If it holds true for $n = k$ and not $n = k+1$, then \[ \frac{\frac{k(k+1)}{2}}{k^2} \ge \frac{b_k}{k^2} > \frac{b_{k+1}}{(k+1)^2} \]so $\frac{(k+1)^3}{2k} > b_{k+1}$. Since we assume the claim does not hold for $n = k+1$, $b_{k+1} \ge \frac{k^2+3k+4}{2}$. Plugging this into $b_{k+1}$ and simplifying yields $k < 1$, contradiction.

Therefore $b_{k+1} \le \frac{(k+1)(k+2)}{2}$, and notice that if the two are equal, then \[ \frac{\frac{k(k+1)}{2}}{k^2} > \frac{\frac{(k+1)(k+2)}{2}}{(k+1)^2}. \]This simplifies to $1 > 0$ which is obviously true.

Since $\frac{b_k}{k^2} \le \frac{\frac{k(k+1)}{2}}{k^2} = \frac{1}{2} + \frac{1}{2k}$, $r$ is at max $\frac{1}{2}$. Now consider $0 \le r < 1/2$.

Claim: $\frac{b_n}{n^2} - r \le \frac{1}{n}$.
Proof: Let $b_{n+1}$ be the maximum integer such that $\frac{b_n}{n^2} > \frac{b_{n+1}}{(n+1)^2}$, so $\frac{b_{n+1} + 1}{(n+1)^2} \ge \frac{b_n}{n^2}$. Rearranging, \[ \frac{b_n}{n^2} - \frac{b_{n+1}}{(n+1)^2} \le \frac{1}{(n+1)^2} < \frac{1}{(n)(n+1)} = \frac{1}{n} - \frac{1}{n+1}. \]Noticing the telescoping sum, we put \[ \frac{b_n}{n^2} - r \le \sum_{p=n}^{\infty} \frac{1}{p} - \frac{1}{p+1} = \frac{1}{n}. \]
Now consider the construction where $b_1 = 1$ and for $k > 1$, $b_k$ is the minimum positive integer value such that $\frac{b_k}{k^2} - r \ge \frac{1}{k}$. Since $\frac{b_k - 1}{k^2} - r < \frac{1}{k}, \frac{b_k}{k^2} - r < \frac{1}{k} + \frac{1}{k^2}$ so if this sequence exists, it converges to $r$.

Also, \[ \left( \frac{b_{k+1}}{(k+1)^2} - \frac{b_k}{k^2} \right) + \left( \frac{b_k}{k^2} - r \right)  \ge \frac{1}{n} - \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{1}{n+1} \]so since a value of $b_{k+1}$ can be found and it must be a positive (bounded from below) integer, we can find a minimum value of $b_{k+1}$. Therefore this creates a valid sequence.

Therefore all values $0 \le r \le \frac{1}{2}$ can be possible infimums and all other values are impossible.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
blueprimes
362 posts
#39
Y by
We claim the answer is $0 \le r \le \dfrac{1}{2}$, the lower bound is obvious.

$\textbf{Claim 1:}$ $r \le \dfrac{1}{2}$ and is precisely the maxima.

We greedily choose $b_k$ according to the recurrence $b_{k + 1} = \left\lfloor \dfrac{(k + 1)^2}{k^2} b_k \right\rfloor$. We will show that in fact, $b_n = \dfrac{n(n + 1)}{2}$ by induction.

The base case is obvious, now assuming up to an arbitrary $n$ to show for $n + 1$ we require
\[ \dfrac{(n + 1)(n + 2)}{2}  = \left \lfloor \dfrac{(n + 1)^3}{2n} \right \rfloor \iff \dfrac{(n + 1)(n + 2)}{2} \le \dfrac{(n + 1)^3}{2n} < \dfrac{(n + 1)(n + 2)}{2} + 1.\]Expanding gives $n^3 + 3n^2 + 2n \le n^3 + 3n^2 + 3n + 1 < n^3 + 3n^2 + 4n$ which is true.

Now $\dfrac{b_n}{n^2} = \dfrac{1}{2} + \dfrac{1}{2n}$ which obviously approaches $\dfrac{1}{2}$ as $n$ grows large.

$\textbf{Claim 2:}$ All values $0 \le r \le \dfrac{1}{2}$ are obtainable.

Consider the sequence $b_n = \dfrac{n(n + 1)}{2}$ for $n \le K - 1$ and $b_n = \lceil rn^2 \rceil + n$ for $n \ge K$ where $K$ is sufficiently large enough such that $\dfrac{b_n}{n^2} > r$. We claim that this construction works. Indeed,
\[ r < \dfrac{\lceil rn^2 \rceil + n}{n^2} < \dfrac{rn^2 + n + 1}{n^2} = r + \dfrac{1}{n} + \dfrac{1}{n^2}\]so $\lim_{n \to \infty} \dfrac{b_n}{n^2} = r$. Moreover,
\[ \dfrac{\lceil rn^2 \rceil + n}{n^2} > \dfrac{\lceil r(n + 1)^2 \rceil + (n + 1)}{(n + 1)^2} \iff (n + 1)^2 \lceil rn^2 \rceil + n^2 + n > n^2 \lceil r(n + 1)^2 \rceil .\]
Attachments:
This post has been edited 1 time. Last edited by blueprimes, May 4, 2025, 2:37 AM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
shendrew7
799 posts
#40
Y by
Our answer is $\boxed{[0,\tfrac 12]}$, constructed with $b_n = 1$ and $b_n = \tfrac{n(n+1)}{2}$, which can be easily shown to be the extremes.

For the values in between, it suffices to find an expression of the form
\[\frac{b_n}{n^2} = \frac{\lceil rn^2+sn+t \rceil}{n^2}\]
for sufficiently high $n$ (where we just set $b_n = \tfrac{n(n+1)}{2}$ before that), which asymptotically approaches $r$ from above. It suffices to have
\[\frac{\lceil rn^2+sn+t \rceil}{n^2} > \frac{\lceil r(n+1)^2+s(n+1)+t \rceil}{(n+1)^2},\]
which suffices to have
\[\frac{rn^2+sn+t}{n^2} > \frac{r(n+1)^2+s(n+1)+t+1}{(n+1)^2}\]\[s(n^2+n) + t(2t+1) > n^2.\]
Setting $s=1$ and $t=0$ works. $\blacksquare$
Z K Y
N Quick Reply
G
H
=
a