Convergent sequence/series proof 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 $0$.
A proof I was shown:
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 $S_0 := 0$,
$$\begin{align*}y_n &= n^{-1} \sum_{k=1}^nk(S_k- S_{k-1})\\
&= n^{-1}\left(nS_n + \sum_{k=1}^{n-1}S_k (k - (k+1)) \right)\\
&= S_n - n^{-1}\sum_{k=1}^{n-1}S_k
\end{align*}$$
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 $n \geq N$
$$\begin{align*}
\left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| &\leq n^{-1} \sum_{k=1}^N |S_k- S_{k-1}| \\
&\qquad+ n^{-1} \sum_{k=N + 1}^n |S_k- S_{k-1}|\\
&= n^{-1} \sum_{k=1}^N |S_k- S_{k-1}| + \epsilon\left(1 - \frac{N}{n}\right)
\end{align*}$$
Indeed by taking the limsup for both sides we see that for any $\epsilon > 0$
$$0 \leq \limsup_{n \rightarrow \infty} \left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| \leq \epsilon$$
By squeeze theorem,
$$\lim_{n \rightarrow \infty} \left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| = \limsup_{n \rightarrow \infty} \left|n^{-1} \sum_{k=1}^n (S_k- S_{k-1})\right| = 0$$
It then follows that $y_n \to S-S=0$ as $n \to \infty$.
QED.
Is this proof correct? If so, could somebody provide some intuition about it. Any insight much appreciated.
 A: It’s not quite correct, but it can be fixed. The first part is correct, but in case you’re not accustomed to summation by parts, I’ll note that you don’t need that technique:
$$\begin{align*}
\sum_{k=1}^nks_k&=\sum_{k=1}^n\sum_{\ell=1}^ks_k=\sum_{\ell=1}^n\sum_{k=\ell}^ns_k\\
&=\sum_{\ell=1}^n(S_n-S_{\ell-1})=nS_n-\sum_{\ell=1}^nS_{\ell-1}\tag{1}\\
&=nS_n-\sum_{k=1}^{n-1}S_k\,,
\end{align*}$$
since $S_0=0$, and that gives us
$$y_n=S_n-n^{-1}\sum\limits_{k=1}^{n-1}S_k\,.\tag{2}$$
The one sequence whose convergence we do know is the sequence of partial sums $S_n$ of the series $\sum_{k\ge 1}s_k$, so expressing the numbers $y_n$ in terms of those partial sums is a natural thing to try, and $(2)$ accomplishes it. Once we get that, it’s clear that we want $S_n-n^{-1}\sum_{k=1}^{n-1}S_k$ to approach $0$ as $n\to\infty$.
We don’t know much about $\sum_{k=1}^{n-1}S_k$, but we do know that when $n$ and $k$ are large, $|S_n-S_k|$ is small, and the first summation in line $(1)$ shows that
$$y_n=n^{-1}\sum_{k=1}^n(S_n-S_k)\,,$$
and if $n$ is large enough, ‘most’ of the terms in the summation ought to be small. Let $\epsilon>0$. There is an $N\in\Bbb N$ such that $|S_n-S_k|<\epsilon$ whenever $n,k\ge N$, so for $n\ge N$ we have
$$\begin{align*}
|y_n|&=\left|n^{-1}\sum_{k=1}^n(S_n-S_k)\right|\\
&\le n^{-1}\sum_{k=1}^N|S_n-S_k|+n^{-1}\sum_{k=N+1}^n|S_n-S_k|\\
&<n^{-1}\sum_{k=1}^N|S_n-S_k|+\left(1-\frac{N}n\right)\epsilon\,.
\end{align*}$$
It follows that $\lim\limits_{n\to\infty}|y_n|\le\epsilon$ for all $\epsilon>0$ and hence that $\lim\limits_{n\to\infty}y_n=0$.
