Let $\{x_n\}$ for $n \geq 1$ be a sequence of positive real numbers and let $s_n = \frac{x_n}{n}$. Let $\{y_n\}$ for $n \geq 1$ be another sequence such that $y_n = n^{-1} \sum_{i = 1}^{n} x_i$. Prove that if $\sum_{j = 1}^{\infty} s_j$ converges, then $\{y_n\}$ converges to $0$.
My attempt:
Assume that $\sum_{j = 1}^{\infty} s_j$ converges. Because it converges we can infer that $\lim_{n \rightarrow \infty} s_n = 0 \implies \lim_{n \rightarrow \infty} \frac{x_n}{n} = 0$. By the definition of limits, for all $\epsilon > 0$, there is some $N \in \mathbb{N}$ so that for $n \geq N$, we have $|s_n| < \epsilon \implies |\frac{x_n}{n}| < \epsilon$.
We need to show that $\lim_{n\to\infty} y_n = y$ for some $y$. So for all $\epsilon_2 > 0$ we need to find some $N_2$ such that for all $n \geq N_2$, we have $|y_n - y| < \epsilon_2$. But $y_n = n^{-1} \sum_{i = 1}^{n} x_i$ so $|n^{-1} \sum_{i = 1}^{n} x_i - y| < \epsilon_2$
Not too sure what to do from here and I am hoping that somebody can point me in the right direction. I don't think my method of using $\epsilon$ and limits is correct and instead think that the problem could involve some manipulation of series? Any assistance is appreciated.
Update:
Let $S_n = \sum_{j=1}^{n} s_j = \sum_{j=1}^n \frac{x_j}{j}$. Because it converges there is some $S$ such that $\lim_{n \rightarrow \infty} S_n = \sum_{j=1}^{\infty} \frac{x_j}{j} = S$.
We can also write:
$$y_n =n^{-1}\sum_{k=1}^n x_k = n^{-1} \sum_{k=1}^nk \frac{x_k}{k} = n^{-1} \sum_{k=1}^nk s_k$$
Using summation by parts, with $S_0 := 0$,
$$y_n = n^{-1} \sum_{k=1}^nk(S_k- S_{k-1})= n^{-1}\left(nS_n + \sum_{k=1}^{n-1}S_k (k - (k+1)) \right) = S_n - n^{-1}\sum_{k=1}^{n-1}S_k$$
We need to show $n^{-1}\sum_{k=1}^{n-1}S_k \to S$ as $n \to \infty$ when $S_n \to S$.
Since $S_n$ converges to $S$, there is some $N \in \mathbb{N}$ such that for $n \geq N$
$$|n^{-1} \sum_{k=1}^n S_k- S_{k-1}| \leq n^{-1} \sum_{k=1}^N |S_k- S_{k-1}| + n^{-1} \sum_{k=N + 1}^n |S_k- S_{k-1}| = ^{-1} \sum_{k=1}^N |S_k- S_{k-1}| + \epsilon(1 - \frac{N}{n})$$
Indeed by taking the limsup for both sides we see that for any $\epsilon > 0$
$$0 \leq \limsup_{n \rightarrow \infty} |n^{-1} \sum_{k=1}^n S_k- S_{k-1}| \leq \epsilon$$
By squeeze theorem, $\lim_{n \rightarrow \infty} |n^{-1} \sum_{k=1}^n S_k- S_{k-1}| = \limsup_{n \rightarrow \infty} |n^{-1} \sum_{k=1}^n S_k- S_{k-1}| = 0$
It then follows that $y_n \to S-S=0$ as $n \to \infty$.
QED.
Credits to user RRL.
