If $\sum_{n\geq 1}\frac{a_n}{n}$ converges, then $\lim_{n\rightarrow \infty}\frac{1}{n}\sum_{k=1}^na_k=0$

I am dealing with the test of the OBM (Brasilian Math Olympiad), University level, 2016, phase 2.

As I've said at this topic (question 2), I hope someone can help me to discuss this test. Thanks for any help.

The question 1 says:

Let be $(a_n)_{n\geq1}$ a real sequence such that $\sum_{n\geq 1}\dfrac{a_n}{n}$ converges. Prove that:

$\lim_{n\rightarrow \infty}\dfrac{1}{n}\sum_{k=1}^na_k=0$.

My attempt:

I've had an ideia if the sequence is positive. Something like that:

Let be $C=\sum_{n\geq 1}\dfrac{a_n}{n}$. The terms of the sum that we'll lead are nearlier of $0$ than these terms, so the limit exists. (Do I need anything else here?)

Take $n=2^m$. We have $\dfrac{a_i}{n}\leq\dfrac{a_i}{i}-\dfrac{1}{2}\dfrac{a_i}{i}-\ldots - \dfrac{1}{2^j}\dfrac{a_i}{i}$ if $\dfrac{1}{n}\leq \dfrac{1}{i}\bigg(1-\dfrac{1}{2}-\ldots - \dfrac{1}{2^j}\bigg)$, it means, $i\leq n\dfrac{1}{2^j}$. So,

$\dfrac{1}{n}\sum_{k=1}^na_k=$ $\dfrac{a_{2^m}}{2^m}+\dfrac{a_{2^m-1}}{2^m}+\ldots + \dfrac{a_1}{2^m}\leq$ $\sum_{k=1}^{2^m}a_k-\sum_{k=1}^{2^{m-1}}a_k-\sum_{k=1}^{2^{m-2}}a_k-\ldots -\sum_{k=1}^{2^0}a_k$

Carrying to infty,

$\lim_{n\rightarrow \infty}\dfrac{1}{n}\sum_{k=1}^na_k\leq C-\dfrac{1}{2}C-\dfrac{1}{4}C-\ldots=0$

As the terms are positive, the limit is zero, as we need prove.

How can I extend if there's negative terms?

I know there are gaps on my thoughts. I thanks for help.

Solution 1:

Let $s_0=0$ and for $n\geq 1$, $$s_n=\sum_{k=1}^{n}\frac{a_k}k$$ Then if $k\geq 1$, $a_k=k(s_k-s_{k-1})$

We have $$\begin{split} \sum_{k=1}^{n}a_k &= \sum_{k=1}^{n} k(s_k-s_{k-1})\\ &= \sum_{k=1}^{n} ks_k- \sum_{k=1}^{n}ks_{k-1}\\ &=\sum_{k=1}^{n} ks_k-\sum_{k=0}^{n-1} (k+1)s_k\\ &= ns_n -\sum_{k=1}^{n-1} s_k \end{split}$$ In other words $$\frac 1 n \sum_{k=1}^{n}a_k = s_n - \frac 1 n \sum_{k=1}^{n-1} s_k$$ Since by assumption, $s_n$ converges to a limit (let's call it $s$), by Cesaro's summation theorem, so does $\frac 1 n \sum_{k=1}^{n-1} s_k$. It follows that $$\lim_{n\rightarrow+\infty}\frac 1 n \sum_{k=1}^{n}a_k =s-s=0$$

Solution 2:

Let $s_n=\sum_{k=1}^{n} \frac {a_k} k$. Then $a_n=n(s_n-s_{n-1})$. Hence $\frac 1 n \sum_{k=1}^{n} a_k=\frac 1 n (-s_1-s_2-\cdots-s_{n-1}+ns_n)$ (after some simplification). Now use the fact that $s_n \to s$ implies $\frac 1 {n-1} (-s_1-s_2-\cdots-s_{n-1}) \to -s$ and $\frac n {n-1} \to 1$.