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Let $a_n$ be a sequence, $S_n=\sum_{k=1}^n a_k$.

(1) If $S_n$ is bounded, $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, show $\lim_{n\to\infty}a_n=0$.

(2). If $\lim_{n\to\infty}\frac{S_n}{n}=0$, $\lim_{n\to\infty} (a_{n+1}-a_n)=0$, can we show $\lim_{n\to\infty}a_n=0$? Prove it if it true, or else give a counterexample.

On the first problem, I have tried Cauchy's criteria, arguing by contradiction...But no solution.

On the second, I have tried $a_n=\ln n$, but...

xldd
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2 Answers2

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Suppose that $a_{n+1}-a_n \to 0$, but the $a_n$ fail to converge to $0$. There is an $\epsilon > 0$ and infinitely many values of $n$ such that $|a_n|>\epsilon$. Let's assume there are infinitely many $n$ such that $a_n > \epsilon$ (the opposite case, that $a_n < -\epsilon$ infinitely often, is similar). Let $k$ be a positive integer. Let $N$ be a value such that (1) $a_N>\epsilon$, and (2) for $n \geq N$, $|a_{n+1}-a_n| < \epsilon/k$ (fill in detail: how do you know that such an $N$ exists?). For $N \leq n \leq N+k$ we have $$ a_n \geq \epsilon-(n-N)\frac{\epsilon}{k} = \epsilon \frac{k+N-n}{k}. $$ Now consider $$ S_{N+k}-S_N = a_{N+1}+a_{N+2}+\dotsb+a_{N+k}, $$ and apply the lower bound. You can show (fill in details here) that $S_{N+k}-S_N$ is unbounded, therefore the $S_n$s are unbounded.

Zach Teitler
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  • How to give such a counterexample then? – xldd Aug 09 '17 at 09:32
  • Nice proof +1. I think this technique can be used for second part also. – Paramanand Singh Aug 09 '17 at 10:16
  • Sorry, I don't know what to do for the second part. I can't even tell if it's true or false. – Zach Teitler Aug 09 '17 at 14:01
  • @RRL It is not accurate I think. since such $N$ depends on $k$... So what happens for the second part? – xldd Aug 10 '17 at 01:19
  • @RRL I have fullfilled the details, so I accept that answer. It comes from a friend, who is studying mathematics. You see in (2), Zach use $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, so by $\epsilon-N$ definition, that $N$ in the answer should depends on the $\epsilon/k$ also. – xldd Aug 10 '17 at 02:23
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By Stolz-Cesaro,

$$\lim_{n \to \infty} \frac{a_n}{n} = \lim_{n \to \infty} (a_{n+1} - a_n) = 0,$$

and if $a_n \to a$,

$$\lim_{n \to \infty} \frac{S_n}{n} = \lim_{n \to \infty}a_n = a $$

Consequently, to find a counterexample for part (2) we need a sequence where $\lim_{n \to \infty}a_n$ does not exist and $|a_n|$ grows no faster than $n$.

A suitable counterexample is $a_n = \sin \sqrt{n}$.

The limit does not exist, since the subsequence $a_{n^2} = \sin n$ is easily shown not to converge. We also have $$|\sin \sqrt{n+1} - \sin \sqrt{n}| \leqslant \sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} \to 0,$$

as well as,

$$\tag{*}\sum_{k=1}^n \sin \sqrt{k} = O(\sqrt{n}),$$

whence,

$$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \sin \sqrt{k} = 0$$

The result (*) can be proved by comparing the sum with $\int_1^n \sin \sqrt{x} \, dx = O(\sqrt{n})$.

RRL
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  • Very instructive counter-example. +1 – Paramanand Singh Aug 11 '17 at 02:19
  • @ParamanandSingh: Thanks. This was a somewhat tricky problem despite the simplicity of the statement, and I have never seen anything similar. Also, its a good example of reverse CS not working where $S_n/n \to 0$ does not imply $a_n \to 0$, even with a stronger hypothesis. – RRL Aug 11 '17 at 04:03