Sufficient conditions to ensure the limit is zero 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...
 A: 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.
A: 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 $\int_1^n \sin \sqrt{x} \, dx = O(\sqrt{n})$. 
