# Arithmetic mean sequence, is this proof correct?

If $$a_n\to a\in\mathbb{R}$$ then $$\frac{a_1+\cdots + a_n}{n}\to a$$

I start by proving the case $$a_n\to 0$$ then I know how to generalize to the case $$a\in\mathbb{R}$$ (my aim here is more to be sure that this type of reasoning below is correct, rather than answering this especific question).

Given $$\epsilon>0$$ choose $$N$$ such that $$n>N \implies |a_n|<\epsilon /2$$. After that $$N$$ chosen, choose $$N_0$$ such that $$N_0 > \frac{2|a_1 + \cdots + a_N|}{\epsilon}$$.

Then $$n > \max\{N, N_0\}$$ implies $$\left|\frac{a_1 + \cdots + a_n}{n}\right| \le \left|\frac{a_1 + \cdots + a_N}{n}\right| + \left|\frac{a_{N+1} + \cdots + a_n}{n}\right| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$$ Then $$\left|\frac{a_1 + \cdots + a_n}{n}\right|\to 0$$

Is this correct? Thanks in advance.

• Your proof for sure is not correct. Sep 22 '19 at 21:05
• @DeepSea can you please explain why? Sep 22 '19 at 21:22
I would put a couple more steps in, but the idea is correct. You have not shown how $$N,N_0$$ come into it. I would write \begin {align}\left|\frac{a_1 + \cdots + a_n}{n}\right| &\le \left|\frac{a_1 + \cdots + a_N}{n}\right| + \left|\frac{a_{N+1} + \cdots + a_n}{n}\right| \\ &\le \left|\frac{a_1 + \cdots + a_N}{N_0}\right| + \left|\frac{(n-N+1)\frac \epsilon 2}{n}\right|\\ &\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} \\&= \epsilon \end {align}