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Prove that if $a_1,a_2,\ldots $ converges to $a$ then $\lim_{n\rightarrow\infty} \dfrac{\sum_{i=1}^n a_i}{n}= a$.

Let $\epsilon>0$. We know that there is natural $N$ such that $\vert a_n-a\vert <\epsilon$ whenever $n>N$. Since $(a_n)$ is convergent we also know that it is bounded by some constant $M>0$.

Now,

$$\left \vert \frac{\sum_{i=1}^n a_i}{n} -a \right\vert \le \frac{\sum_{i=1}^N \vert a_i-a\vert }{n} + \frac{\sum_{i=N+1}^n \vert a_i-a\vert }{n} \le \frac{NM}{n}+\frac{n-N-1}{n} \epsilon \rightarrow \epsilon.$$

Since $\epsilon$ was any positive number we must conclude that $\left\vert \dfrac{\sum_{i=1}^n a_i}{n} -a \right \vert$ converges to $0$.

Is this a correct explanation? How can I make it more "explicit"?

Edit (thanks to Kitty): Obviously $M$ must be a bound for $|a_i-a|$, $i=1,\ldots,N$.

luka5z
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    I'm fine with that. You might google for Cesáro mean for more information – Hagen von Eitzen Jan 05 '16 at 18:10
  • Thank you sir!! I will. I was just a bit doubtful about my argument "Since $\epsilon$ was any positive number we must conclude that". It didnt look rigorous enough for me. – luka5z Jan 05 '16 at 18:13
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    @luka5z That is fine. You've shown the limit of $0\le | \frac{1}{n} \sum_{i=1}^n a_i - a|$ is at most $\epsilon$, for any positive number $\epsilon$, so the limit must be zero. – angryavian Jan 05 '16 at 18:15
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    One error: the $M$ in your proof is not the bound for $(a_i)$. It is $\max_{i\leq N} {|a_i-a|}$. – KittyL Jan 05 '16 at 18:16
  • @angryavian So I basically sandwiched it? – luka5z Jan 05 '16 at 18:17
  • @KittyL you are right, I missed that. Thank you. – luka5z Jan 05 '16 at 18:17

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