$a:=\lim_{n\to\infty}\sum_{k=1}^n\frac{a_k}{n}$, $k|a_{k+1}-a_k|<1, $prove $\lim_{n\to\infty}a_n=a$

Question

Let $\{a_n\}$ be a sequence of real numbers.
If $a:=\lim_{n\to\infty}\sum_{k=1}^n\frac{a_k}{n}$ exists and $\forall{k}\in\mathbb{N^+}\quad k|a_{k+1}-a_k|<1$.

Please prove $\lim_{n\to\infty}a_n=a.$

The limit of its average do not guarantee the existance of the sequence's limit. Also the second one do not. But together they do.
I have no idea about this one. Hint will also be appreciated.
Thanks in advance.

Edit: Maybe it's not enough to use only one term of its average. Here is my effort: $|\frac{1}{n+l}(\sum_{k=1}^{n+l}{a_k})-a_n|\\= \frac{1}{n+l}|\sum_{k=1}^{n-1}{(a_k-a_n)}+\sum_{k=n+1}^{n+l}{(a_k-a_n)}|\\\leq \frac{1}{n+l}|\sum_{k=1}^{n-1}{(a_n-a_k)}|+\frac{1}{n+l}|\sum_{k=n+1}^{n+l}{(a_k-a_n)}|\\= \frac{1}{n+l}|\sum_{k=1}^{n-1}{k(a_{k+1}-a_k)}|+\frac{1}{n+l}|\sum_{k=n+1}^{n+l-1}{(n+l-k)(a_{k+1}-a_k)}|\\\leq \frac{1}{n+l}\sum_{k=1}^{n-1}|{k(a_{k+1}-a_k)}|+\frac{1}{n+l}\sum_{k=n+1}^{n+l-1}|{(n+l-k)(a_{k+1}-a_k)}|\\\leq \frac{n-1}{n+l}+\frac{1}{n+l}\sum_{k=n+1}^{n+l-1}\frac{n+l-k}{k}\\= \frac{n-1}{n+l}+\frac{1}{n+l}\sum_{k=1}^{l-1}\frac{k}{n+l-k}\\= \frac{n-1}{n+l}+\sum_{k=1}^{l-1}(\frac{1}{n+l-k}-\frac{1}{n+l})\\\leq \frac{n-l}{n+l}+\ln(\frac{n+l}{n})\stackrel{min}{\longrightarrow}{\ln 2}$

Answer 1

Set $a_0 = 0$ and note that

$$\frac{1}{n}\sum_{k=1}^na_k = \frac{1}{n}\sum_{k=1}^n\sum_{j=1}^k(a_j-a_{j-1}) = \frac{1}{n}\sum_{j=1}^n\sum_{k=j}^n(a_j-a_{j-1}) = \frac{1}{n}\sum_{j=1}^n(n- j+1)(a_j-a_{j-1}) \\ =a_n - \frac{1}{n}\sum_{j=1}^n(j-1)(a_j-a_{j-1})$$

Hence,

$$\tag{*}\left|a_n - a\right| \leqslant \left|\frac{1}{n}\sum_{k=1}^n a_k - a \right|+ \frac{1}{n} \sum_{j=1}^n (j-1)|a_j - a_{j-1}|$$

If we have the stronger condition $k|a_{k+1}-a_k| \to 0$ as $k \to \infty$ rather than $k|a_{k+1} - a_k| < 1$, then by Stolz-Cesaro

$$\tag{**}\lim_{n \to \infty}\frac{1}{n} \sum_{j=1}^n (j-1)|a_j - a_{j-1}| = \lim_{n\to \infty} n|a_{n+1}-a_n| = 0$$

Using (*) and (**) we get $\lim_{n \to \infty} a_n = a.$

Under the condition: $k|a_{k+1} - a_k| < 1$

The theorem is also true when $k|a_{k+1} - a_k| = \mathcal{O}(1)$. Apparently this result was proved by Hardy.

The first step would be, with $m < n$,

$$a_n - \frac{1}{n}\sum_{k=1}^na_k = \frac{1}{n-m}\sum_{k=m+1}^n (a_n - a_k) +\frac{1}{n-m}\sum_{k=m+1}^n a_k - \frac{1}{n}\sum_{k=1}^na_k \\ = \frac{1}{n-m}\sum_{k=m+1}^n (a_n - a_k) + \frac{m}{n-m} \left(\frac{1}{n} \sum_{k=1}^n a_k -\frac{1}{m} \sum_{k=1}^m a_k \right)$$

The rest of the details are discussed here.