Convergence of a sequence $\lim_{n\to\infty}a_n=0$ Let $a_n$ be a sequence, $s_n=\sum_{i=1}^n a_i$ be partial sum.
(1) If $s_n$ is bounded, $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, show that $\lim_{n\to\infty}a_n=0$.
(2) If $\lim_{n\to\infty}s_n/n=0$, $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, do we have the statement $\lim_{n\to\infty}a_n=0$? If so, prove it, if not, construct a counterexample.
The first statement, I have no idea, just $a_n=a_1+\sum_{i=1}^{n-1}(a_{i+1}-a_i)$.
The second statement is wrong, I think; however, I could not provide a counterexample.
 A: Theorem:
If $\;\left\{a_n\right\}_{n\in\mathbb{N}}\;$ is a sequence such that $\;\lim_\limits{n\to\infty}\left(a_{n+1}-a_n\right)=0\;$ and there exists $\;M\ge0\;$ such that $\;\left|\sum_\limits{i=1}^n a_i\right|\le M\;$ for all  $\;n\in\mathbb{N}\;,\;\;$ then $\;\lim_\limits{n\to\infty}a_n=0\;.$
Proof:
Let $\;s_n=\sum_\limits{i=1}^n a_i\;$ for all $\;n\in\mathbb{N}\;.$
So, for hypothesis, it results that $\;\left|s_n\right|\le M\;$ for all $\;n\in\mathbb{N}\;.$
It means that the sequence $\;\left\{s_n\right\}_{n\in\mathbb{N}}\;$ of the partial sums is bounded.
Now we are going to prove this theorem through a proof by contradiction.
If the thesis were false, there would exist $\;\epsilon_0>0\;$ for which
$\forall n\in\mathbb{N}\;\;\exists k_n\in\mathbb{N},\; k_n>n\;$ such that $\;\left|a_{k_n}\right|\ge\epsilon_0\;.$
Thus we get a sequence $\left\{k_n\right\}_{n\in\mathbb{N}}$ of positive integers such that
$k_n>n\;$ and $\;\left|a_{k_n}\right|\ge\epsilon_0\;\;$ for all $\;n\in\mathbb{N}\;.$
Let $\;q=\lfloor\frac{4M}{\epsilon_0}\rfloor+1\in\mathbb{N}\;.$
Since $\;\lim_\limits{n\to\infty}\left(a_{n+1}-a_n\right)=0\;$, there exists $\;\nu\in\mathbb{N}\;$ such that
$\left|a_{n+1}-a_n\right|<\frac{\epsilon_0}{q+1}\;\;$ for all $\;n>\nu\;.\color{blue}{\quad(*)}$
Seeing as $\;k_\nu>\nu\;,\;$ from $\;(*)\;$ for $\;k_\nu\le n\le k_\nu+q-1\;$ we get that
$-\frac{\epsilon_0}{q+1}<a_{k_\nu+1}-a_{k_\nu}<\frac{\epsilon_0}{q+1}\;,$
$-\frac{\epsilon_0}{q+1}<a_{k_\nu+2}-a_{k_\nu+1}<\frac{\epsilon_0}{q+1}\;,$
$-\frac{\epsilon_0}{q+1}<a_{k_\nu+3}-a_{k_\nu+2}<\frac{\epsilon_0}{q+1}\;,$
. . . . . . . . . . . . . . . . . . . . . . . . . .
$-\frac{\epsilon_0}{q+1}<a_{k_\nu+q}-a_{k_\nu+q-1}<\frac{\epsilon_0}{q+1}\;.$
By adding the first two of the previous inequalities, then the first three, the first four, etc, we also get that
$-\frac{\epsilon_0}{q+1}<a_{k_\nu+1}-a_{k_\nu}<\frac{\epsilon_0}{q+1}\;,$
$-2\frac{\epsilon_0}{q+1}<a_{k_\nu+2}-a_{k_\nu}<2\frac{\epsilon_0}{q+1}\;,$
$-3\frac{\epsilon_0}{q+1}<a_{k_\nu+3}-a_{k_\nu}<3\frac{\epsilon_0}{q+1}\;,$
. . . . . . . . . . . . . . . . . . . . . . . . . .
$-q \frac{\epsilon_0}{q+1}<a_{k_\nu+q}-a_{k_\nu}<q\frac{\epsilon_0}{q+1}\;.$
And, by adding the last inequalities, we get that
$\left|s_{k_\nu+q}-s_{k_\nu}-q a_{k_\nu}\right|<\frac{q\left(q+1\right)}{2}\frac{\epsilon_0}{q+1}=\frac{q\epsilon_0}{2}\;.$
Moreover,
$\left|s_{k_\nu+q}-s_{k_\nu}\right|\ge$
$\ge\left|q a_{k_\nu}\right|-\left|s_{k_\nu+q}-s_{k_\nu}-q a_{k_\nu}\right|>$
$>q\left|a_{k_\nu}\right|-\frac{q\epsilon_0}{2}\ge$
$\ge q\epsilon_0-\frac{q\epsilon_0}{2}=\frac{1}{2}q\epsilon_0=$
$=\frac{1}{2}\left(\lfloor\frac{4M}{\epsilon_0}\rfloor+1\right)\epsilon_0>\frac{1}{2}\frac{4M}{\epsilon_0}\epsilon_0=2M\;.$
So it results that
$\left|s_{k_\nu+q}-s_{k_\nu}\right|>2M\;.\color{blue}{\quad(**)}$
But, since $\;\left\{s_n\right\}_{n\in\mathbb{N}}\;$ is bounded for hypothesis, it follows that
$\left|s_{k_\nu+q}-s_{k_\nu}\right|\le\left|s_{k_\nu+q}\right|+\left|s_{k_\nu}\right|\le M+M=2M\;,$
which contradicts the inequality $\;(**)$.
So the thesis cannot be false otherwise it would lead to a contradiction.
Hence the thesis is true and it means that
$\lim_\limits{n\to\infty}a_n=0\;.$
A: For any $N\gt0$, we have
$$\begin{align}
s_{n+N}-s_n
&=a_{n+N}+a_{n+N-1}+\cdots+a_{n+1}\\
&=(a_{n+N}-a_{n+N-1})+2(a_{n+N-1}-a_{n+N-2})+\cdots+N(a_{n+1}-a_n)+Na_n
\end{align}$$
Now if $s_n$ is bounded, then there is some $M\gt0$ such that $|s_n|\le M$ for all $n$.  It follows that
$$|a_n|\le{2M\over N}+|a_{n+N}-a_{n+N-1}|+|a_{n+N-1}-a_{n+N-2}|+\cdots+|a_{n+1}-a_n|$$
Given $\epsilon\gt0$, choose $N$ so that $2M/N\lt\epsilon/2$ and note that, since $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, there is an $m$ such that $|a_{k+1}-a_k|\lt\epsilon/(2N)$ for all $k\ge m$. With these choices of $N$ and $m$, we have
$$|a_n|\lt{\epsilon\over2}+{\epsilon\over2N}+{\epsilon\over2N}+\cdots+{\epsilon\over2N}={\epsilon\over2}+{\epsilon\over2}=\epsilon$$
for all $n\gt m$, and thus, since $\epsilon$ can be arbitrarily small, $\lim_{n\to\infty}a_n=0$.
Added later: As for part (2), consider the following sequence for $a_1,a_2,a_3,\ldots$:
$$0,0,{1\over2},0,{-1\over2},0,{1\over3},{2\over3},{1\over3},0,{-1\over3},{-2\over3},{-1\over3},0,0,0,0,0,0,0,{1\over4},{2\over4},{3\over4},{4\over4},{3\over4},{2\over4},{1\over4},0,{-1\over4},{-2\over4},{-3\over4},{-4\over4},{-3\over4},{-2\over4},{-1\over4},0,\ldots,0,{1\over5},\ldots{},{8\over5},\ldots,{-8\over5},\ldots,{-1\over5},0,\ldots$$
where $a_n=0$ except for a stretch after each index $n=k!$ which goes up in increments of $1/k$ to $2^{k-2}/k$, back down to $-2^{k-2}/k$, and then back up to $0$. (Note, each stretch is of length $2^k-1$, which is less than $(k+1)!-k!$, so there is always room for it, with a buffer of $0$'s between stretches.) It's clear that $a_{n-1}-a_n\to0$ as $n\to\infty$, not hard to see that $s_n/n\to0$, and easy to see that $a_n\not\to0$ as $n\to\infty$ (indeed, the sequence $a_n$ is unbounded, since $2^{k-2}/k\to\infty$ as $k\to\infty$).
A: Here goes the proof of (1):
Suppose that $\lim_{n\to\infty}a_n=0$ doesn't hold, then, there exists $\varepsilon_0>0$ and $N\in \mathbb{N}$ such that for all $n\geq N$, $|a_n|\geq \varepsilon_0$. Here, we distinguish two cases:
CASE I: There are infinitely many negative and non-negative terms.
Then, by the condition $\lim_{n\to\infty}(a_{n+1}-a_n)=0$, there exists $N_0$ such that for all $n\geq N_0$: $|a_{n+1}-a_n|<\varepsilon_0$, then pick some $n\geq \max\{{N_0,N}\}$ such that $a_n$ and $a_{n+1}$ have different sign, then $|a_{n+1}-a_n|\geq 2\varepsilon_0$ and $|a_{n+1}-a_n|< \varepsilon_0$, which is absurd.
CASE II: Case I doesn't hold.
Then, w.l.o.g., suppose that $M\in \mathbb{N}$ is such that $a_n$ is non-negative for all $n\geq M$, then, for all $n\geq \max\{{M,N}\}$: $a_n\geq \varepsilon_0$. This directly implies that $s_n$ can not be bounded.
This contradiction shows that $\lim_{n\to\infty}a_n=0$.
About (2), I had the idea $a_n= 1+1/n^2$, for this sequence we have $$\lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2}\right)=0 \ \ ,$$
$$\frac{s_n}{n} = \frac{n+\sum^n_1 \frac{1}{k^2}}{n}$$
The last expression doesn't go to zero, but, it's a nice try, maybe you can fix it.
A: Theorem 1:
$\left|\sum\limits_{h=1}^n\sin\sqrt{h}\right|<3\sqrt{n}+2\;\;,\quad\forall n\in\mathbb{N}\;.$
Proof:
We apply the following Euler-Maclaurin summation formula
$\sum\limits_{a<h\le b} f(h)=\int_a^b\left[f(x)+\left(x-\lfloor x\rfloor\right)f’(x)\right]dx$
to the function $\;f(x)=\sin\sqrt{x}\;$.
$\sum\limits_{h=1}^n \sin\sqrt{h}=\int_0^n\left[\sin\sqrt{x}+\left(x-\lfloor x\rfloor\right)\frac{\cos\sqrt{x}}{2\sqrt{x}}\right]dx\;$.
$\int_0^n\sin\sqrt{x}dx=\left[2\sin\sqrt{x}-2\sqrt{x}\cos\sqrt{x}\right]_0^n=\\=2\sin\sqrt{n}-2\sqrt{n}\cos\sqrt{n}\;.$
$\left|\int_0^n\sin\sqrt{x}dx\right|=\left|2\sin\sqrt{n}-2\sqrt{n}\cos\sqrt{n}\right|\le2+2\sqrt{n}\;$.
Since $\;\left|\cos\sqrt{x}\right|\le1\;$ and $\;0\le x-\lfloor x\rfloor<1\;$, we get that
$\left|\left(x-\lfloor x\rfloor\right)\frac{\cos\sqrt{x}}{2\sqrt{x}}\right|\le\frac{x-\lfloor x\rfloor}{2\sqrt{x}}<\frac{1}{2\sqrt{x}}\;,$
for all $\;x\in\left]0,+\infty\right[\;$.
Hence,
$\left|\int_0^n\left(x-\lfloor x\rfloor\right)\frac{\cos\sqrt{x}}{2\sqrt{x}}dx\right|\le\int_0^n\left|\left(x-\lfloor x\rfloor\right)\frac{\cos\sqrt{x}}{2\sqrt{x}}\right|dx<\\<\int_0^n\frac{1}{2\sqrt{x}}dx=\sqrt{n}\;.$
Therefore,
$\left|\sum\limits_{h=1}^n\sin\sqrt{h}\right|\le\left|\int_0^n\sin\sqrt{x}dx\right|+\left|\int_0^n\left(x-\lfloor x\rfloor\right)\frac{\cos\sqrt{x}}{2\sqrt{x}}dx\right|<$
$<2+2\sqrt{n}+\sqrt{n}=3\sqrt{n}+2\;,$
for all $\;n\in\mathbb{N}\;$.

Theorem 2:
The sequence $\left\{a_n\right\}_{n\in\mathbb{N}}=\left\{\sin\sqrt{n}\right\}_{n\in\mathbb{N}}$ satisfies the following properties:

*

*$\lim\limits_{n\to\infty}\cfrac{s_n}{n}=\lim\limits_{n\to\infty}\cfrac{\sum\limits_{h=1}^n a_h}{n}=0\;;$


*$\lim\limits_{n\to\infty} \left(a_{n+1}-a_n\right)=0\;;$


*$\nexists\;\lim\limits_{n\to\infty}a_n\;.$
Proof:
Since $\;\left|\cfrac{s_n}{n}\right|=\left|\cfrac{\sum\limits_{h=1}^n\sin\sqrt{h}}{n}\right|<\cfrac{3\sqrt{n}+2}{n}\;,$ for all $\;n\in\mathbb{N}\;,$ and $\;\lim\limits_{n\to\infty}\cfrac{3\sqrt{n}+2}{n}=0\;,$
it follows that $\;\lim\limits_{n\to\infty}\cfrac{s_n}{n}=0\;.$
Moreover,
$\lim\limits_{n\to\infty}\left(a_{n+1}-a_n\right)=\lim\limits_{n\to\infty}\left(\sin\sqrt{n+1}-\sin\sqrt{n}\right)=\\=\lim\limits_{n\to\infty}\left[2\cos\left(\frac{\sqrt{n+1}+\sqrt{n}}{2}\right)\sin\left(\frac{1}{2\sqrt{n+1}+2\sqrt{n}}\right)\right]=0\;,$
indeed it is a limit of a product of a bounded sequence and an infinitesimal sequence.
If there existed the limit $\;\lim\limits_{n\to\infty} a_n\;,$ this limit would be finite, indeed the sequence $\;\left\{a_n\right\}_{n\in\mathbb{N}}\;$ is bounded, consequently $\;\lim\limits_{n\to\infty} a_n=l\in\mathbb{R}\;$.
And, by applying a theorem about subsequences, we get that
$\lim\limits_{n\to\infty}a_{n^2}=\lim\limits_{n\to\infty}\sin n=l\;,$
but it contradicts the fact that $\;\lim\limits_{n\to\infty}\sin n\;$ does not exist.
Hence the limit $\;\lim\limits_{n\to\infty}a_n\;$ cannot exist, otherwise it would lead to a contradiction.
So we have proved by reductio ad absurdum that the limit $\;\lim\limits_{n\to\infty}a_n\;$ does not exist.

Theorem 3:
If $\;\left\{a_n\right\}_{n\in\mathbb{N}}\;$ is a sequence such that $\;\lim\limits_{n\to\infty}\cfrac{s_n}{n}=\lim\limits_{n\to\infty}\cfrac{\sum\limits_{h=1}^n a_h}{n}=0\;$ and there exists the limit $\;\lim\limits_{n\to\infty}a_n\;,\;$ then $\;\lim\limits_{n\to\infty} a_n=0\;.$
Proof:
Since there exists $\;\lim\limits_{n\to\infty}a_n\;,\;$ by applying a corollary of Stolz-Cesaro Theorem, we get that there also exists the limit $\;\lim\limits_{n\to\infty}\cfrac{s_n}{n}\;$ and $\;\lim\limits_{n\to\infty}\cfrac{s_n}{n}=\lim\limits_{n\to\infty}a_n\;.$
But, from hypothesis, we know that $\;\lim\limits_{n\to\infty}\cfrac{s_n}{n}=0\;.$
Hence, by the uniqueness theorem for limits, it follows that
$\lim\limits_{n\to\infty}a_n=0\;.$
