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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.

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  • $\begingroup$ Your proof for sure is not correct. $\endgroup$
    – DeepSea
    Sep 22 '19 at 21:05
  • $\begingroup$ @DeepSea can you please explain why? $\endgroup$ Sep 22 '19 at 21:22
  • $\begingroup$ See also this question and the ones linked to it. $\endgroup$
    – Arnaud D.
    Sep 22 '19 at 21:58
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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 $$\begin {align}\left|\frac{a_1 + \cdots + a_n}{n}\right| &\le \left|\frac{a_1 + \cdots + a_N}{n}\right| + \left|\frac{a_{N+1} + \cdots + a_n}{n}\right| \\ &\le \left|\frac{a_1 + \cdots + a_N}{N_0}\right| + \left|\frac{(n-N+1)\frac \epsilon 2}{n}\right|\\ &\lt \frac{\epsilon}{2} + \frac{\epsilon}{2} \\&= \epsilon \end {align}$$

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