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Suppos $a_n$ is a real sequence and $a_n \to a$. Define $s_n = \dfrac {a_1 + \cdots + a_n}{n}$. Show that $s_n$ converges to $a$.

I know that this problem has been solved on this website many times, but I wanted to come up with a proof of my own and ask if my proof is correct. If yes, how could it be improved?

My proof:

Fix $\epsilon>0$. There exists some $N$ such that $n \ge N \implies |a_n-a| < \epsilon$. Take any $n \ge N$. We have $$|s_n-a| = \left| \dfrac {a_1 + \cdots + a_n}{n} - a \right| = \dfrac {1}{n} \left|\sum_{k=1}^n (a_k-a) \right| \le \dfrac {1}{n} \sum_{k=1}^n |a_k-a|$$

Now let $A = \sum_{k=1}^{N-1} |a_k-a|$. Then the above is equal to

$$\dfrac {1}{n} \left( A + \sum_{k=N}^n |a_k-a| \right) \le \dfrac An + \dfrac {n-N}{n} \cdot \epsilon \le \dfrac An + \dfrac {n}{n} \cdot \epsilon \le \dfrac An + \epsilon$$ Now there exists an $M \ge N$ such that $n \ge M \implies \dfrac An < \epsilon$. Therefore for $n \ge M$, we have $|s_n-a|< 2 \epsilon$ and we are done.

joriki
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    Look good to me. – copper.hat Dec 11 '19 at 04:29
  • @copper.hat Thanks! Was somewhat nervous since all the other proofs I've seen seems longer and/or more complicated than this one. – Blue Dec 11 '19 at 04:31
  • For reference, this is known as the Cesàro mean of the sequence $a_n$. – Math1000 Dec 11 '19 at 05:38
  • A minor point: You should end up with $\lt\epsilon$, not $\lt2\epsilon$, so you should start with $n \ge N \implies |a_n-a| < \frac12\epsilon$. – joriki Dec 11 '19 at 05:43
  • @joriki it's sloppy but it's not wrong to conclude that if something is $<2\epsilon$ for any $\epsilon > 0$ then it is $0$. – Jürgen Sukumaran Dec 11 '19 at 13:30
  • @TSF: There's no conclusion here that something is zero. The task is to show that given $\epsilon\gt0$ we can find $M$ such that for all $n\ge M$ we have $|s_n-a|\lt\epsilon$. Of course it's equivalent to show that given $\epsilon\gt0$ we can find $M$ such that for all $n\ge M$ we have $|s_n-a|\lt2\epsilon$; but it's one more implicit step that the reader has to perform in their mind, so it's not ideal. As I wrote, it's just a minor point. – joriki Dec 11 '19 at 13:48
  • You are concluding that $|s_n-a|$ is $0$ in the limit. Sorry for poor phrasing. The idea is the same as what I said originally, that $2\epsilon$ is just as arbitrary, for lack of a better word, as $\epsilon$. – Jürgen Sukumaran Dec 11 '19 at 13:52
  • I know you wanted to know if your proof was correct, but it doesn't seem to be very different from the one I linked. It's also pretty much the same as this this answer for what it's worth. – Arnaud D. Dec 11 '19 at 15:12

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