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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
jlacosta
Mar 2, 2025
0 replies
A problem
kwin   0
a few seconds ago
Let x, y, z > 0 and $ x^2+y^2+z^2=3 $ . Find max:
$$P=xy(x^2+y^2-\frac{1}{z})+yz(y^2+z^2-\frac{1}{x})+zx(x^2+z^2-\frac{1}{y})$$
0 replies
kwin
a few seconds ago
0 replies
tangent line to 3 circles of center A, B, C and all pass through orthocenter
parmenides51   1
N 3 minutes ago by ilovemath0402
Source: 2018 Saudi Arabia BMO TST I p4
Let $ABC$ be an acute, non isosceles with $I$ is its incenter. Denote $D, E$ as tangent points of $(I)$ on $AB,AC$, respectively. The median segments respect to vertex $A$ of triangles $ABE$ and $ACD$ meet$ (I)$ at$ P,Q,$ respectively. Take points $M, N$ on the line $DE$ such that $AM \perp BE$ and $AN \perp  C D$ respectively.
a) Prove that $A$ lies on the radical axis of $(MIP)$ and $(NIQ)$.
b) Suppose that the orthocenter $H$ of triangle $ABC$ lies on $(I)$. Prove that there exists a line which is tangent to three circles of center $A, B, C$ and all pass through $H$.
1 reply
parmenides51
Jul 25, 2020
ilovemath0402
3 minutes ago
geometry party
pnf   0
6 minutes ago
pnf
6 minutes ago
0 replies
n Tans
MithsApprentice   17
N 26 minutes ago by AshAuktober
Source: USAMO 1998
Let $a_0,a_1,\cdots ,a_n$ be numbers from the interval $(0,\pi/2)$ such that \[ \tan (a_0-\frac{\pi}{4})+ \tan (a_1-\frac{\pi}{4})+\cdots +\tan (a_n-\frac{\pi}{4})\geq n-1.  \] Prove that \[ \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}.  \]
17 replies
MithsApprentice
Oct 9, 2005
AshAuktober
26 minutes ago
Differentiation Marathon!
LawofCosine   181
N Today at 12:24 AM by Levieee
Hello, everybody!

This is a differentiation marathon. It is just like an ordinary marathon, where you can post problems and provide solutions to the problem posted by the previous user. You can only post differentiation problems (not including integration and differential equations) and please don't make it too hard!

Have fun!

(Sorry about the bad english)
181 replies
LawofCosine
Feb 1, 2025
Levieee
Today at 12:24 AM
Integrals problems and inequality
tkd23112006   2
N Yesterday at 6:02 PM by PolyaPal
Let f be a continuous function on [0,1] such that f(x) ≥ 0 for all x ∈[0,1] and
$\int_x^1 f(t) dt \geq \frac{1-x^2}{2}$ , ∀x∈[0,1].
Prove that:
$\int_0^1 (f(x))^{2021} dx \geq \int_0^1 x^{2020} f(x) dx$
2 replies
tkd23112006
Feb 16, 2025
PolyaPal
Yesterday at 6:02 PM
real analysis
ay19bme   1
N Yesterday at 5:30 PM by alexheinis
.........
1 reply
ay19bme
Yesterday at 3:05 PM
alexheinis
Yesterday at 5:30 PM
Proving an inequality involving cosine functions
pii-oner   3
N Yesterday at 3:33 PM by pii-oner
Hello AoPS Community,

I am curious about how to demonstrate the following inequality:

[code] \sqrt{1 - |\cos(x \pm y)|^a} \leq \sqrt{1 - |\cos(x)|^a} + \sqrt{1 - |\cos(y)|^a}, \quad \text{for } a \geq 1. [/code]

I’ve plotted the functions, and the inequality seems to hold. However, I am looking for a rigorous way to prove it.

I’d greatly appreciate insights into how to break this down analytically and references to similar problems or techniques that might be helpful.

Thank you so much for your guidance and support!
3 replies
pii-oner
Jan 22, 2025
pii-oner
Yesterday at 3:33 PM
Linear algebra
Dynic   2
N Yesterday at 3:32 PM by loup blanc
Let A and B be two square matrices with the same size. Prove that if AB is an invertible matrix, then A and B are also invertible matrices
2 replies
Dynic
Yesterday at 2:48 PM
loup blanc
Yesterday at 3:32 PM
Very hard group theory problem
mathscrazy   3
N Yesterday at 9:50 AM by quasar_lord
Source: STEMS 2025 Category C6
Let $G$ be a finite abelian group. There is a magic box $T$. At any point, an element of $G$ may be added to the box and all elements belonging to the subgroup (of $G$) generated by the elements currently inside $T$ are moved from outside $T$ to inside (unless they are already inside). Initially $
T$ contains only the group identity, $1_G$. Alice and Bob take turns moving an element from outside $T$ to inside it. Alice moves first. Whoever cannot make a move loses. Find all $G$ for which Bob has a winning strategy.
3 replies
mathscrazy
Dec 29, 2024
quasar_lord
Yesterday at 9:50 AM
limit of u(pi/45)
EthanWYX2009   0
Yesterday at 7:08 AM
Source: 2025 Pi Day Challenge T5
Let \(\omega\) be a positive real number. Divide the positive real axis into intervals \([0, \omega)\), \([\omega, 2\omega)\), \([2\omega, 3\omega)\), \([3\omega, 4\omega)\), \(\ldots\), and color them alternately black and white. Consider the function \(u(x)\) satisfying the following differential equations:
\[
u''(x) + 9^2u(x) = 0, \quad \text{for } x \text{ in black intervals},
\]\[
u''(x) + 63^2u(x) = 0, \quad \text{for } x \text{ in white intervals},
\]with the initial conditions:
\[
u(0) = 1, \quad u'(0) = 1,
\]and the continuity conditions:
\[
u(x) \text{ and } u'(x) \text{ are continuous functions}.
\]Show that
\[
\lim_{\omega \to 0} u\left(\frac{\pi}{45}\right) = 0.
\]
0 replies
EthanWYX2009
Yesterday at 7:08 AM
0 replies
Integration Bee Kaizo
Calcul8er   42
N Yesterday at 12:57 AM by awzhang10
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
42 replies
Calcul8er
Mar 2, 2025
awzhang10
Yesterday at 12:57 AM
maximum value
Tip_pay   3
N Friday at 9:37 PM by alexheinis
Find the value $x\in [0,4]$ at which the function $f(x)=\int_{0}^{\sqrt{x}}\ln \frac{e}{1+t^2}dt$ takes its maximum value
3 replies
Tip_pay
Friday at 8:48 PM
alexheinis
Friday at 9:37 PM
Spheres and a point source of light
mofidy   3
N Friday at 9:22 PM by kiyoras_2001
How many spheres are needed to shield a point source of light?
Unfortunately, I didn't find a suitable solution for it on the page below:
https://artofproblemsolving.com/community/c4h1469498p8521602
Here too, two different solutions are given:
https://math.stackexchange.com/questions/2791186
3 replies
mofidy
Mar 11, 2025
kiyoras_2001
Friday at 9:22 PM
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
1276 posts
#5 • 2 Y
Y by Adventure10, Ritwin
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
688 posts
#6 • 1 Y
Y by Adventure10
Solution
This post has been edited 5 times. Last edited by Shaddoll, Apr 30, 2017, 9:46 PM
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william122
1576 posts
#7 • 1 Y
Y by Adventure10
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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#8
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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
3783 posts
#9
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Solution
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asdf334
7577 posts
#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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#11
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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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#12
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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
7577 posts
#13
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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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