If the positive numbers $v_n$ approach zero as $n\to\infty$, prove that their average $(v_1+v_2+\dots+v_n)/n$ also approaches zero.
We have:
For any $\epsilon$, there is an $N$ such that $v_n<\epsilon$ if $n>N$.
We need to prove:
For any $\epsilon$, there is an $N$ such that $\frac{v_1+v_2+\dots+v_n}n<\epsilon$ if $n>N$.
I am having a hard time connecting these two statements. I have tried with examples but I have not been able to get through these symbols. Intuitively, average follows instantaneous. $\bar v$ rises and falls with the next $v$. But I need a formal proof.
As you have observed, for any $\epsilon$, there is an $N$ such that $v_n<\epsilon$ if $n>N$.
So \begin{align*} \frac{v_1+v_2+\dots+v_n}{n}&=\frac{v_1+v_2+\dots+v_N+v_{N+1}+\dots+v_n}{n}\\ &<\frac{v_1+v_2+\dots+v_N+\epsilon+\dots+\epsilon}{n}\\ &=\frac{v_1+v_2+\dots+v_N}{n}+\frac{(n-N)}{n}\epsilon \end{align*}
It should be clear that $\frac{v_1+v_2+\dots+v_N}{n}\to 0$.
Also, $\frac{(n-N)}{n}\epsilon\to \epsilon$.
So overall, $\lim\frac{v_1+v_2+\dots+v_n}{n}\leq\epsilon$. Since $\epsilon>0$ is arbitrary, it tends to zero.