Suppose $\lbrace a_n \rbrace$ is a real sequence and define $s_n=\frac{1}{n}\sum_{k=1}^{n}a_k$
Prove that if $\lim_{n\to\infty}a_n=0$, then $\lim_{n\to\infty}s_n=0$
I find it hard to go this way; I know if that if the $\frac{1}{n}$ wasn't in that term, it wouldn't necessarily be true, but I'm not sure how to take advantage of its introduction to actually evaluate the limit of $s_n$. What's the clearest step-by-step way to show this?
Since $\lim\limits_{n\to \infty}\,a_{n} = 0$ we have
$\forall \varepsilon>0 \,\, \exists N>0$ such that $|{a_{n} - 0}|<\varepsilon \quad \forall n > N$
Using this, we have
$|s_{n} - 0| = |\frac{1}{n}\sum\limits_{k=1}^{n}a_{k}| = |\frac{1}{n}\sum\limits_{k=1}^{N}a_{k}+\frac{1}{n}\sum\limits_{N+1}^{n}a_{k}| \leq |\frac{1}{n}\sum\limits_{k=1}^{N}a_{k}| + |\frac{1}{n}\sum\limits_{N+1}^{n}a_{k}|$
$< |\frac{1}{n}\cdot constant| + |\frac{1}{n}\cdot(n-N)\cdot\varepsilon|$
When $n\to\infty \, , \, |s_{n}-0|<\varepsilon$