How to show that $\lim_{n\to \infty} \frac{a_1 +a_2 + \cdots + a_n}{n} = 0?$

Question

We are given that $\displaystyle\sum_{k=1}^\infty \frac{a_k}{k}$ converges, and we want to show that $$\lim_{n\to \infty} \frac{a_1 +a_2 + \cdots + a_n}{n} = 0.$$ Let $\epsilon>0.$ Then since $\displaystyle\sum_{k=1}^\infty \frac{a_k}{k}$ converges we have that there is some positive integer $N_\epsilon$ such that $\left|\sum_{k=N_\epsilon}^\infty \dfrac{a_k}{k}\right|<\dfrac{\epsilon}2$. Let $\sum_{k=1}^{N_\epsilon} a_k =A$ and choose $N_1$ so that $\left|\frac{A}{n}\right|<\frac{\epsilon}{2}$ when $n\geq N_1$. Now let $n\geq \max \left\{N_1, N_\epsilon \right\}$ so that $$\left|\frac{a_1 +a_2 + \cdots + a_n}{n}\right|\leq\left|\frac{A}{n}\right|+\left|\frac{\sum_{k=N_\epsilon+1}^n a_k}{n}\right|<\left|\frac{A}{n}\right|+ \left|\sum_{k=N_\epsilon+1}^\infty \frac{a_k}{k}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$ Again, I do not feel convinced of my own argument. Am I going the wrong direction? Is there perhaps a better route to showing my conclusion? Any help is appreciated.

EDIT: Don’t know why this question got closed. I’m not solely wondering how to prove this statement, I’m also wondering if my proof is sufficient.

Answer 1

I'll prove with your ideas.

Given that $\sum_{k=1}^{\infty}\frac {a_{k}}k$converges, for each $\epsilon$, there exist an $N_{\epsilon}$, such that$$|\sum_{k=N_\epsilon}^{\infty}\frac {a_k}k|<\frac \epsilon2$$for arbitrary $N'>N_\epsilon$, we have $$|\sum_{k=N_\epsilon}^{N'}\frac {a_k}{N'}|<|\sum_{k=N_\epsilon}^{N'}\frac {a_k}k|<\frac \epsilon2$$ while fix this $\epsilon$, $$\forall N_1>\frac {|2{\sum_{k=1}^{N_{}\epsilon}}{a_k}|}{\epsilon},|\frac {\sum_{k=1}^{N_{\epsilon}}{a_k}}{N_1}|<\frac \epsilon2$$ So, set N as max{$N_1,N_{\epsilon,}$}, we can get the conclusion$$\forall n>N,|\frac {\sum_{k=1}^{n}a_n}n|\leq |\frac {\sum_{k=1}^{N_{}\epsilon}{a_k}}n|+|\frac {\sum_{k=N_{}\epsilon}^{n}{a_k}}n|\leq \epsilon$$as the select of $\epsilon$ is arbitrary, the proof is complete.

Answer 2

For a "better route", use summation by parts with $S_0 = 0$ and $S_k = \sum_{j=1}^k \frac{a_j}{j}$ for $k > 0$ to get

$$\sum_{k=1}^n a_k = \sum_{k=1}^n k\frac{a_k}{k} = \sum_{k=1}^n k(S_k - S_{k-1}) = nS_n - \sum_{k=1}^{n-1}S_k(k+1 - k), $$

and, thus,

$$\frac{1}{n} \sum_{k=1}^n a_k = S_n - \frac{1}{n} \sum_{k=1}^{n-1}S_k$$

Finish by showing that the limit of the RHS is $0$. That $S_n \to S$ implies $n^{-1}\sum_{k=1}^n S_k \to S$ is a well known result.