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.


marked as duplicate by Daniel Fischer Dec 13 '16 at 14:20

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  • $\begingroup$ See here. I think we have better versions, but I haven't yet found them. $\endgroup$ – Daniel Fischer Dec 13 '16 at 14:08
  • $\begingroup$ It might be a bit confusing to use the same $\epsilon$ and $N$ in your assumption and in your required result. What happens if you take 2N in what you want to prove? what happens if you take 3N? or in general if you take kN for some integer k? $\endgroup$ – Ofir Dec 13 '16 at 14:13

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.


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