In one of my classes we showed that the sequence of Cesàro sums for a convergent sequence $(a_n)$ is convergent directly. My professor mentioned you could also show convergence by the Cauchy criterion, so now I want to prove that the Cesàro sum sequence is Cauchy myself. My proof thus far looks like this:
We want to show that for all $\epsilon > 0$, there exists $N\in \mathbb{N}$ such that: $$\bigg| \frac{a_1+\ldots+a_n}{n} - \frac{a_1+\ldots+a_m}{m} \bigg| < \epsilon $$ for all $n > m \geq N$, where $n > m$ by assumption. Since $(a_n)$ converges, $(a_n)$ is Cauchy, in particular $\forall\epsilon > 0$, $\exists N\in\mathbb{N}$ such that $|a_n - a_m| < \epsilon$ for $n > m \geq N$. Now, we can rearrange the left-hand side of the inequality we want to show like so: $$\bigg| \frac{a_1+\ldots+a_n}{n} - \frac{a_1+\ldots+a_m}{m} \bigg|$$ $$\bigg| \frac{m}{mn}(a_1+\ldots+a_n) - \frac{n}{mn}(a_1+\ldots+a_m) \bigg|$$ $$\bigg| \frac{(m-n)}{mn}(a_{m+1}+\ldots+a_n) \bigg|$$ $$\bigg| \frac{(m-n)}{mn} \bigg| \bigg| (a_{m+1}+\ldots+a_n) \bigg|$$ $$\bigg| \frac{(m-n)}{mn} \bigg| \bigg| a_n - a_m \bigg|$$ So the inequality in what we want to show is just a multiple of the inequality for the sequence we know is Cauchy. Here's where I get stuck. I can verify that a constant times a Cauchy sequence is still Cauchy, but does that apply here? Is the fraction in the end step even a constant if it varies with every $m,n$? I don't think so. Where can I go from here? Is this even the right direction?