Let $\sigma_n = \frac{1}{n}(s_1 + \ldots + s_n)$. Assuming that $\lim_{n \rightarrow \infty} s_n = s$, prove that $\lim_{n \rightarrow \infty} \sigma_n = s$.
How do I start with this proof? It makes intuitive sense, but I don't know how to do it formally. I've tried writing down the definition of the limit but have not gotten very far, so some hints would help. Thanks!
This falls into a standard pattern where there are bad parts and good parts.
In this case, for any $\epsilon > 0$, there is a $N(\epsilon)$ such that $|s_n - s| < \epsilon$ for $n \ge N(\epsilon)$. This is the good part.
The bad part has the other $n$, for which you might have $|s_n - s| \ge \epsilon$.
Write
$\begin{array}\\ S(n) &=\dfrac{\sum_{k=1}^{n} s_k}{n}\\ &=\dfrac{\sum_{k=1}^{N(\epsilon)-1} s_k}{n} +\dfrac{\sum_{k=N(\epsilon)}^{n} s_k}{n}\\ \text{so that}\\ S(n)-s &=\dfrac{\sum_{k=1}^{n} s_k}{n}-s\\ &=\dfrac{\sum_{k=1}^{n} (s_k-s)}{n}\\ &=\dfrac{\sum_{k=1}^{N(\epsilon)-1} (s_k-s)}{n} +\dfrac{\sum_{k=N(\epsilon)}^{n} (s_k-s)}{n}\\ &=D_1+D_2\\ \end{array} $
We want to show that $S(n)-s$ can be made as small as we want.
$D_1$ has the bad part, where the $s_k$ might be far from $s$.
$D_2$ has the good part, where all the $s_k$ are close to $s$. For this, we immediately have $|D_2| < \epsilon$ since each $|s_k-s| < \epsilon$.
If we can make $|D_1| < \epsilon$, then $S(n) - s| < 2\epsilon$, and we are done. Note that it doesn't matter that we have $2\epsilon$ instead of $\epsilon$.
We can do this by making $n$ large.
Let $d(\epsilon) =\max_{k=1}^{N(\epsilon)-1} |s_k-s| $. Then
$\begin{array}\\ |D_1| &=\big|\dfrac{\sum_{k=1}^{N(\epsilon)-1} (s_k-s)}{n}\big|\\ &\le\dfrac{\sum_{k=1}^{N(\epsilon)-1} |s_k-s|}{n}\\ &\le\dfrac{\sum_{k=1}^{N(\epsilon)-1} d(\epsilon)}{n}\\ &\le\dfrac{N(\epsilon) d(\epsilon)}{n}\\ \end{array} $
Therefore, if $\dfrac{N(\epsilon) d(\epsilon)}{n} < \epsilon$, or $n > \dfrac{N(\epsilon) d(\epsilon)}{\epsilon} $, $|D_1| < \epsilon$, which is want we want.
None of this is original. This is standard, and a good thing to know.
