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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
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jlacosta
Mar 2, 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
Inequality => square
Rushil   11
N 3 minutes ago by lksb
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
11 replies
Rushil
Oct 7, 2005
lksb
3 minutes ago
p + q + r + s = 9 and p^2 + q^2 + r^2 + s^2 = 21
who   28
N 14 minutes ago by asdf334
Source: IMO Shortlist 2005 problem A3
Four real numbers $ p$, $ q$, $ r$, $ s$ satisfy $ p+q+r+s = 9$ and $ p^{2}+q^{2}+r^{2}+s^{2}= 21$. Prove that there exists a permutation $ \left(a,b,c,d\right)$ of $ \left(p,q,r,s\right)$ such that $ ab-cd \geq 2$.
28 replies
who
Jul 8, 2006
asdf334
14 minutes ago
H not needed
dchenmathcounts   44
N an hour ago by Ilikeminecraft
Source: USEMO 2019/1
Let $ABCD$ be a cyclic quadrilateral. A circle centered at $O$ passes through $B$ and $D$ and meets lines $BA$ and $BC$ again at points $E$ and $F$ (distinct from $A,B,C$). Let $H$ denote the orthocenter of triangle $DEF.$ Prove that if lines $AC,$ $DO,$ $EF$ are concurrent, then triangle $ABC$ and $EHF$ are similar.

Robin Son
44 replies
dchenmathcounts
May 23, 2020
Ilikeminecraft
an hour ago
IZHO 2017 Functional equations
user01   51
N an hour ago by lksb
Source: IZHO 2017 Day 1 Problem 2
Find all functions $f:R \rightarrow R$ such that $$(x+y^2)f(yf(x))=xyf(y^2+f(x))$$, where $x,y \in \mathbb{R}$
51 replies
user01
Jan 14, 2017
lksb
an hour ago
chat gpt
fuv870   2
N an hour ago by fuv870
The chat gpt alreadly knows how to solve the problem of IMO USAMO and AMC?
2 replies
fuv870
an hour ago
fuv870
an hour ago
Inequality with wx + xy + yz + zw = 1
Fermat -Euler   23
N an hour ago by hgomamogh
Source: IMO ShortList 1990, Problem 24 (THA 2)
Let $ w, x, y, z$ are non-negative reals such that $ wx + xy + yz + zw = 1$.
Show that $ \frac {w^3}{x + y + z} + \frac {x^3}{w + y + z} + \frac {y^3}{w + x + z} + \frac {z^3}{w + x + y}\geq \frac {1}{3}$.
23 replies
Fermat -Euler
Nov 2, 2005
hgomamogh
an hour ago
Waiting for a dm saying me again "old geometry"
drmzjoseph   0
2 hours ago
Source: Idk easy
Given $ABCD$ a tangencial quadrilateral that is not a rhombus, let $a,b,c,d$ be lengths of tangents from $A,B,C,D$ to the incircle of the quadrilateral which center is $I$. Let $M,N$ be the midpoints of $AC,BD$ resp. Prove that
\[ \frac{MI}{IN}=\frac{a+c}{b+d} \]
0 replies
drmzjoseph
2 hours ago
0 replies
Finally hard NT on UKR MO from NT master
mshtand1   2
N 2 hours ago by IAmTheHazard
Source: Ukrainian Mathematical Olympiad 2025. Day 1, Problem 11.4
A pair of positive integer numbers \((a, b)\) is given. It turns out that for every positive integer number \(n\), for which the numbers \((n - a)(n + b)\) and \(n^2 - ab\) are positive, they have the same number of divisors. Is it necessarily true that \(a = b\)?

Proposed by Oleksii Masalitin
2 replies
mshtand1
Mar 13, 2025
IAmTheHazard
2 hours ago
IMOC 2017 G5 (<A=120 => E, F, Y,Z are concyclic, incenter related)
parmenides51   4
N 2 hours ago by ehuseyinyigit
Source: https://artofproblemsolving.com/community/c6h1740077p11309077
We have $\vartriangle ABC$ with $I$ as its incenter. Let $D$ be the intersection of $AI$ and $BC$ and define $E, F$ in a similar way. Furthermore, let $Y = CI \cap DE, Z = BI \cap DF$. Prove that if $\angle BAC = 120^o$, then $E, F, Y,Z$ are concyclic.
IMAGE
4 replies
parmenides51
Mar 20, 2020
ehuseyinyigit
2 hours ago
Bosnia and Herzegovina JBMO TST 2013 Problem 1
gobathegreat   3
N 3 hours ago by DensSv
Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2013
It is given $n$ positive integers. Product of any one of them with sum of remaining numbers increased by $1$ is divisible with sum of all $n$ numbers. Prove that sum of squares of all $n$ numbers is divisible with sum of all $n$ numbers
3 replies
gobathegreat
Sep 16, 2018
DensSv
3 hours ago
D1015 : A strange EF for polynomials
Dattier   0
3 hours ago
Source: les dattes à Dattier
Find all $P \in \mathbb R[x,y]$ with $P \not\in \mathbb R[x] \cup \mathbb R[y]$ and $\forall g,f$ homeomorphismes of $\mathbb R$, $P(f,g)$ is an homoemorphisme too.
0 replies
Dattier
3 hours ago
0 replies
P, Q,R collinear and U, R, O, V concyclic wanted, cyclic ABCD, circumcenters
parmenides51   2
N 3 hours ago by DensSv
Source: 2012 Romania JBMO TST2 P4
The quadrilateral $ABCD$ is inscribed in a circle centered at $O$, and $\{P\} = AC \cap BD, \{Q\} = AB \cap CD$. Let $R$ be the second intersection point of the circumcircles of the triangles $ABP$ and $CDP$.
a) Prove that the points $P, Q$, and $R$ are collinear.
b) If $U$ and $V$ are the circumcenters of the triangles $ABP$, and $CDP$, respectively, prove that the points $U, R, O, V$ are concyclic.
2 replies
parmenides51
May 29, 2020
DensSv
3 hours ago
Unsolved Diophantine(I think)
Nuran2010   1
N 3 hours ago by Nuran2010
Find all solutions for the equation $2^n=p+3^p$ where $n$ is a positive integer and $p$ is a prime.(Don't get mad at me,I've used the search function and did not see a correct and complete solution anywhere.)
1 reply
Nuran2010
Mar 14, 2025
Nuran2010
3 hours ago
2^a + 3^b + 1 = 6^c
togrulhamidli2011   1
N 3 hours ago by CM1910
Find all positive integers (a, b, c) such that:

\[
2^a + 3^b + 1 = 6^c
\]
1 reply
togrulhamidli2011
Today at 12:34 PM
CM1910
3 hours ago
Calculus rather than inequalities
darij grinberg   12
N Yesterday at 8:39 PM by asdf334
Source: German TST, IMO ShortList 2003, algebra problem 3
Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]For any pair define $c_n = \min\{a_i,b_i\}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$.


(1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded?

(2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$?

Justify your answer.
12 replies
darij grinberg
Jul 15, 2004
asdf334
Yesterday at 8:39 PM
Calculus rather than inequalities
G H J
Source: German TST, IMO ShortList 2003, algebra problem 3
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darij grinberg
6555 posts
#1 • 3 Y
Y by Adventure10, Mango247, and 1 other user
Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]For any pair define $c_n = \min\{a_i,b_i\}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$.


(1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded?

(2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$?

Justify your answer.
This post has been edited 2 times. Last edited by djmathman, May 27, 2018, 3:38 PM
Reason: overhauled problem wording to fit https://anhngq.files.wordpress.com/2010/07/imo-2003-shortlist.pdf
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jmerry
12096 posts
#2 • 6 Y
Y by Adventure10, Mango247, ehuseyinyigit, and 3 other users
I'm surprised this was a high school contest problem.

For convenience, let the sequences be nonincreasing rather than strictly decreasing- we can always tweak them after the fact without changing the convergence of the series.
(a) Yes. An example:
Let $r_0=0$, $r_1=1$, $r_2=1+2^1$, $r_3=r_2+2^3$, ..., $r_{n+1}=r_n+2^{\frac{n(n+1)}{2}}$.
Let $a_1=1$, $a_k=2^{-3}$ if $r_1<k \le r_3$, $a_k=2^{-m(2m+1)}$ if $r_{2m-1}< k \le r_{2m+1}$.
Let $b_k=2^{-1}$ if $0<k \le r_2$, $b_k=2^{-m(2m-1)}$ if $r_{2m-2}< k \le r_{2m}$.
Then $c_k=2^{-\frac{n(n+1)}{2}}$ if $r_{n-1}<k \le r_n$ and
$\displaystyle \sum_{k=1}^{\infty}c_k = \sum_{j=0}^{\infty} \sum_{k=r_j+1}^{r_{j+1}} 2^{-\frac{(j+1)(j+2)}{2}} = \sum_{j=0}^{\infty} \sum_{k=r_j+1}^{r_{j+1}} 2^{-\frac{j(j+1)}{2}}\cdot 2^{-(j+1)} = \sum_{j=0}^{\infty} 2^{-(j+1)} = 1$

On the other hand $\displaystyle \sum_{k=1}^{r_{2n-1}}a_k > \sum_{j=0}^{n-1} \sum_{k=r_{2j}+1}^{r_{2j+1}}a_k = \sum_{j=0}^{n-1} \sum_{k=r_{2j}+1}^{r_{2j+1}}2^{-j(2j+1)} = \sum_{j=0}^{n-1} 1 = n$
and
$\displaystyle \sum_{k=1}^{r_{2n}}b_k > \sum_{j=1}^n \sum_{k=r_{2j-1}+1}^{r_{2j}}b_k = \sum_{j=1}^{n} \sum_{k=r_{2j-1}+1}^{r_{2j}}2^{-j(2j-1)} = \sum_{j=1}^n 1 = n$
Each series $\sum_k a_k$ and $\sum_k b_k$ diverge, while $\sum_k \min(a_k,b_k)$ converges.

(b)The answer changes: $\sum_k c_k$ must diverge.
Case 1. Suppose that for all $k \ge N$, $a_k < \frac{1}{k}$.
Then $\displaystyle \sum_{k=N}^{\infty} c_k = \sum_{k=N}^{\infty} a_k$ diverges by divergence of $\sum_k a_k$.
Case 2. There exist infinitely many $m$ such that $a_m \ge \frac{1}{m}$
Let $m_n$ be an infinite sequence with $m_{n+1} \ge 2m_n$ and $a_{m_n} \ge \frac{1}{m_n}$ for all $n \ge 1$ (with $m_0=0$). Then $c_{m_n}=\frac{1}{m_n}$, so
$\displaystyle \sum_{k=1}^{m_n} c_k = \sum_{j=0}^{n-1} \sum_{k=m_j+1}^{m_{j+1}} c_k \ge \sum_{j=0}^{n-1} \sum_{k=m_j+1}^{m_{j+1}} \frac{1}{m_{j+1}} \ge \sum_{j=0}^{n-1}\frac{m_{j+1}-m_j}{m_{j+1}} \ge \sum_{j=0}^{n-1} \frac{1}{2} = \frac{n}{2}$
and $\sum_k c_k$ diverges.
These two case exhaust all possibilities, and we are done.
This post has been edited 2 times. Last edited by darij grinberg, Oct 27, 2018, 8:22 PM
Reason: latex fix
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jgnr
1343 posts
#3 • 2 Y
Y by yojan_sushi, Adventure10
I think I've got another construction for (a), but it's somewhat difficult to explain.

$(a_k)=\boxed{\frac12,\frac12},\frac18,\frac1{16},\frac1{32},\frac1{64},\boxed{\underbrace{\frac1{64},\frac1{64},\ldots,\frac1{64}}_{64times}},\frac1{2^{71}},\frac1{2^{72}},\ldots$

${(b_k)=\frac12,\frac14,\boxed{\frac14,\frac14,\frac14,\frac14},\frac1{128},\frac1{256},\ldots,\frac1{2^{70}},\boxed{\underbrace{\frac1{2^{70}},\frac1{2^{70}},\ldots,\frac1{2^{70}}}_{2^{70}times}}},\ldots$

The boxed blocks are terms whose sum is 1, so $(A_k)$ and $(B_k)$ are clearly unbounded. Since $c_i=\frac1{2^i}$, then $(C_k)$ is bounded.
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va2010
1276 posts
#4 • 3 Y
Y by Adventure10, Mango247, Ritwin
I'd like to point out that fedja posted a beautiful solution here
fedja wrote:
This question is asked now and then. The answer is that the series with the maxima in the denominators can converge despite both initial series diverge and the reason is exactly as you put it: ak and bk can overtake each other infinitely many times and accumulate large sums when at rest from the work as maxima. Formalizing it isn't hard and did did an excellent job in this respect. The question is whether we can make it "obvious", i.e., to see it in our heads without touching pen or paper at all. Of course, we can talk about decreasing sequences ak,bk>0 and consider ∑kmin(ak,bk). Now just imagine two people walking. The condition is that neither of them is allowed to increase his speed at any time and that the slowest one carries a stick, which is magically teleported to the other person when he becomes slower. The question is whether both people can walk to infinity while the stick will travel only finite distance on foot. Now the strategy should become clear. Fix any upper bound v for speeds and any upper bound d for the distance the stick is allowed to travel on foot. Let the first person walk at speed v until he goes the distance 1. The second person should crawl slowly (but steadily) all that time at the speed dv/2, so he travels only distance d/2 with the stick. At that moment, the first person slows down enormously to the speed d2v/4, so while the second person continues to go at the speed dv/2 and moves distance 1, the first person (who now has the stick) goes only d/2. By the end of this cycle, both people moved by at least 1 but the stick traveled only distance d on foot. We also end up with the new bound for the speed, which is vd2/4. Now repeat the cycles making the allowed distances for the stick smaller and smaller so that the corresponding series converges. We do not care how fast the speed bounds decay because no matter what the bound is, we still can cover the unit distance in some finite time. During each cycle, each person moves by distance 1 or more, so both ultimately walk away. However, the stick goes only finite distance on foot.

Also, see NIMO Winter Contest 2013.6 for an equivalent problem.
This post has been edited 1 time. Last edited by va2010, Apr 1, 2015, 5:06 PM
Reason: added links to NIMO problem
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va2010
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#5 • 2 Y
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The main idea behind a good deal of these constructions is that the monotonically decreasing condition can be dropped. Indeed, this is because we can divide terms into a bunch of very close but decreasing terms. Then fedja's construction becomes clear: take $d = 1.5$ and take the sums:

$1 + \frac{d}{4} + 1 + \frac{d}{16} \cdots$ which obviously diverges and
$\frac{d}{2} + 1 + \frac{d}{8} + 1 \cdots$ which also diverges. But when you take the minimum, you get

$\frac{d}{2} + \frac{d}{4} + \frac{d}{8} + \frac{d}{16} \cdots = d$, which is convergent. A very nice problem, and an even nicer solution!
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Shaddoll
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#6 • 1 Y
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Solution
This post has been edited 5 times. Last edited by Shaddoll, Apr 30, 2017, 9:46 PM
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william122
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#7 • 1 Y
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1) Yes. We define $\{c_i\}=\frac{1}{2^i}$, and partition $\{a_i\}$ and $\{b_i\}$ into C-runs, which are contiguous blocks of terms equal to the corresponding terms in $\{c_i\}$, and leeks, which are everything else. Note that the C-runs in $\{a_i\}$ and $\{b_i\}$ are disjoint, and perfectly cover $\{c_i\}$. Now, we recursively define $a,b$ as follows: After a C-run which ends with $\frac{1}{2^i}$, construct a leek which consists of $2^i$ copies of $\frac{1}{2^i}$ while the other sequence has its own C-run. The first few terms, for example, are as follows: $$\{a_i\}=\underline{\frac{1}{2}},\frac{1}{2},\frac{1}{2},\underline{\frac{1}{16},\frac{1}{32},\frac{1}{64},\frac{1}{128},\frac{1}{256},\cdots}\\$$$$\{b_i\}=\frac{1}{2},\underline{\frac{1}{4},\frac{1}{8}},\hspace{0.18cm}\frac{1}{8},\hspace{0.18cm}\frac{1}{8},\hspace{0.18cm}\frac{1}{8},\hspace{0.3cm}\frac{1}{8},\hspace{0.3cm}\frac{1}{8}\cdots$$where the C-runs are underlined. This way, $A_n,B_n$ are both unbounded, since each leek adds $1$ to the cumulative sum, however $C_n$ is bounded by 1.

2) The answer is no. Note that $\{b_i\}$ must have infinitely many C-runs, or else $A_i$ and $C_i$ will have the same convergence. Now, define the sequence $\{x_i\}$ to be the indices of the ends of all the C-runs in $\{b_i\}$. Note that $c_{i}\ge \frac{1}{x_j}\forall 1\le i\le x_j$, so $C_{x_j}-C_{x_k}\ge\frac{x_j-x_k}{x_j}=1-\frac{x_k}{x_j}$. As the $\{x_i\}$ are unbounded, construct a subsequence $\{y_i\}$ such that $y_n\ge 2y_{n-1}$. Now, we have $$C_{y_n}\ge \sum_{i=2}^n C_{y_i}-C_{y_{i-1}}\ge \sum_{i=2}^n 1-\frac{y_{n-1}}{y_n}\ge \frac{n-1}{2}$$So, $C_i$ is unbounded.
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everythingpi3141592
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Could someone tell what is the motivation behind the construction in the first part? I tried to construct by taking $a_1 = 1$ and $b_1 = 100$, and continually reducing $b$ by $1$ and keeping $a$ constant untill we reach $(1, 1)$. And then, we further reduce $b$, untill we repeat the same procedure, but by continually reducing $a$ by some number untill again, $b$ become $100$ times $a$. What general strategies did people use to find the powers of $2$ solution?
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vsamc
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#9
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Solution
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asdf334
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#10
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solved very nice but i read the solution and stuff at first so im gonna come back and review in the future
First we solve the first part. The idea is to choose $\{c_n\}:=\left\{2^{-1},2^{-2},\dots,\right\}$ and then choose
\[\{a_n\}:=\left\{2^{-1},2^{-1},2^{-3},2^{-4},2^{-5},2^{-5},\dots,2^{-5},\dots,\right\}\]\[\{b_n\}:=\left\{2^{-1},2^{-2},2^{-2},2^{-2},2^{-2},2^{-6},\dots,2^{-36},\dots,\right\}\]in which both $\{a_n\}$ and $\{b_n\}$ contain infinitely many runs of identical fractions summing to $1$ and therefore have divergent $A_n$ and $B_n$.
Now we prove that the second part is impossible. Consider the set $S$ of indices $i$ where $a_i\ge b_i$. Define $T$ to be the set of indices $i$ where $a_i\le b_i$. The sum of all $a_i$ for $i\in T$ is convergent due to convergence of $C_n$. Hence the sum of all $a_i$ for $i\in S$ is divergent.
Split the indices of $S$ into runs of consecutive values $[d_i+1,d_i+r_i]$. If $d_1=0$ then ignore the first run; it will not affect the solution. For $j\in [d_i+1,d_i+r_i]$ we have
\[a_j\le a_{d_i}\le b_{d_i}=\frac{1}{d_i}.\]
Now notice
\[\frac{1}{d_i+1}+\dots+\frac{1}{d_i+r_i}=\ln(d_i+r_i)-\ln(d_i)+O(1)\]is convergent, thus $\frac{d_i+r_i}{d_i}\le M$ for some constant $M$.
Now notice
\[M\sum_{j=1}^{r_i}\frac{1}{d_i+j}\ge \frac{d_i+r_i}{d_i}\sum_{j=1}^{r_i}\frac{1}{d_i+j}\ge \sum_{j=1}^{r_i}\frac{1}{d_i}\ge \sum_{j=1}^{r_i}a_{d_i+j}\]and summing over all $i$ yields a convergent value on the LHS and a divergent value on the RHS, a contradiction. $\blacksquare$
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awesomeming327.
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Define the following sequence recursively: $s_1=1$, $s_n=s_{n-1}+2^{s_{n-1}}$ for all $n\ge 2$. Define $t_n$ to be the nonnegative integer $k$ such that $s_{k}<t_n\le s_{k+1}$ where $s_0$ is defined as $0$. Then, define
\[a_i=\begin{cases}
\frac{1}{2^n}, & \text{if $t_n$ odd}\\
\frac{1}{2^{s_{t_n}}}, & \text{if $t_n$ even}
\end{cases}\]and define
\[b_i=\begin{cases}
\frac{1}{2^n}, & \text{if $t_n$ even}\\
\frac{1}{2^{s_{t_n}}}, & \text{if $t_n$ odd}
\end{cases}\]Note that $s_{t_n}<n$ so $c_n=2^-n$ for all $n$. Thus, $C_n\le 1$ for all $n$. However,
\[A_{s_{2k}}\ge \sum_{j=1}^{k}{\sum_{t_n=2j}{\frac{1}{2^{s_{t_{n}}}}}}\]The number of values $n$ such that $t_n=2j$ is $s_{2j+1}-s_{2j}=2^{s_{t_n}}$, therefore
\[\sum_{t_n=2j}{\frac{1}{2^{s_{t_{n}}}}}= 1\]so $A_{s_{2k}}\ge k$ is unbounded. Similarly, $B$ is unbounded.

$~$
Redefine $s$ so that $a_n<b_n$ if and only if $t_n$ is odd. If $t$ is bounded then $C$ has the same convergence as either $A$ or $B$. If $t$ is unbounded,
\[C_{s_k}\ge \sum_{j=1}^{k}{\frac{s_{k+1}-s_k}{s_{k+1}+1}}=k-\sum_{j=1}^{k}{\frac{s_k+1}{s_{k+1}+1}}\]and it is easy to see why this is unbounded.
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pie854
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darij grinberg wrote:
Consider pairs of the sequences of positive real numbers \[a_1\geq a_2\geq a_3\geq\cdots,\qquad b_1\geq b_2\geq b_3\geq\cdots\]and the sums \[A_n = a_1 + \cdots + a_n,\quad B_n = b_1 + \cdots + b_n;\qquad n = 1,2,\ldots.\]For any pair define ${\color{red}c_n = \min\{a_i,b_i\}}$ and $C_n = c_1 + \cdots + c_n$, $n=1,2,\ldots$.


(1) Does there exist a pair $(a_i)_{i\geq 1}$, $(b_i)_{i\geq 1}$ such that the sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are unbounded while the sequence $(C_n)_{n\geq 1}$ is bounded?

(2) Does the answer to question (1) change by assuming additionally that $b_i = 1/i$, $i=1,2,\ldots$?

Justify your answer.

I think there's a typo. Should be $c_i$ in the colored text.
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asdf334
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oops i should have gotten this its ok i guess
Part 1.
The answer is yes. Consider the following:
\[\{A_n\}=\left\{\frac{1}{2^0}, c_2,c_3,\frac{1}{2^3}, \frac{1}{2^3},\frac{1}{2^3},\frac{1}{2^3},\frac{1}{2^3}, \frac{1}{2^3},\frac{1}{2^3},\frac{1}{2^3},\dots\right\}\]\[\{B_n\}=\left\{c_1, \frac{1}{2^1},\frac{1}{2^1},c_4,c_5,c_6,c_7,c_8,c_9,c_{10},c_{11},\dots\right\}\]where $c_n=\frac{1}{2^n}$. Three things hold: we have $c_n=\min\{a_n,b_n\}$, we have $a_1\ge a_2\ge \dots$ and $b_1\ge b_2\ge \dots$, and the sums of $\{A_n\}$ and $\{B_n\}$ are divergent as each set contains infinitely many consecutive groups of identical terms summing to $1$.
Part 2.
The answer is no. (This makes sense intuitively, since the convergence of sequence $C$ in the first part was allowed because the sequences $A$ and $B$ decreased so quickly; here, sequence $B$ decreases at quite a slow rate.)

..so how to start? The sequence $A$ must have some values smaller than those in sequence $B$, and some values larger. We'd like to create a contradiction; that is, show that sequence $C$ has infinite sum.

How do we show that sequence $C$ has infinite sum? The easiest way is the one that has been used in the first part already: to find infinitely many long subsequences of sequence $C$, each with a finite sum, so that the total of all the subsequences is infinite.

How do we do that? Let's say that we've found several of these subsequences already, and our "pointer" (the next element that is past all of the previous subsequences) is at $c_N$. We'd like to find an element $c_M$ so that $c_N+\dots+c_M$ is a finite, "predictable" (doesn't keep decreasing, for example, since an infinite geometric series wouldn't give us the desired contradiction) sum.

So how do we find such an element $c_M$? Let's first create an inequality: $c_N+\dots+c_M\ge \ell c_M$, where $\ell=M-N+1$. What's the other information we have? Well, there are infinitely many $c_M=\frac{1}{M}$, since otherwise the sum of all $c_M$ for large $M$ becomes equal to the sum of all $a_M$, which isn't allowed.

Small point to mention: why do we choose $c_M=\frac{1}{M}=b_M$? Well, this is the only information we've been given, and it should intuitively yield something important as the denominator of $M$ is a linear variable, in a sense, just like $\ell$. If we try to choose a $c_M=a_M$, in addition, we end up with $c_M\le \frac{1}{M}$, which is actually "worse" to use compared to $c_M=\frac{1}{M}$, since we're trying to make $\ell c_M$ as large as possible.

Okay, so now we just need to show that, for a fixed $N$, there exists some large $M$ where the resultant $\ell c_M$ sum (and the ones after it) will all combine together and diverge. We have:
\[\ell c_M=\frac{M-N+1}{M}\]and by choosing $M>>N$ we can make this close to $1$. That solves the problem, as we've just found a subsequence with sum equal to $1-\epsilon$ and can do this process infinitely, so that the sum of sequence $C$ is divergent. Nice :) unfortunately wasn't able to solve this but typing it out like this helps I think coolio time to lock in $\blacksquare$
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