# 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.

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 $$\lim_{n\rightarrow+\infty}\frac 1 n \sum_{k=1}^{n}a_k =s-s=0$$

• I did not know this Theorem, Stefan, and your link helped a lot. I thank you very much. Nov 8, 2018 at 17:14
• You're welcome! Nov 8, 2018 at 20:10
• (+1) Nicely written. Jul 3, 2021 at 4:16
• Why there are $n-1$ partial sums instead of $n$ as it's written at the page you sent? Jan 7, 2022 at 13:41
• $\frac 1 n \sum_{k=1}^{n-1} s_k=-\frac{s_n}n + \frac 1 n \sum_{k=1}^{n} s_k$ , and $\frac{s_n}n\rightarrow 0$, so it doesn't matter if the sum goes to $n$ or $n-1$. the Jan 7, 2022 at 16:05

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 $$\frac n {n-1} \to 1$$.

• Thanks very much. Your answer is very clear ... congratulations!! Nov 3, 2018 at 23:54