This was an excercise in my exam and I was wondering whether my solution was correct or not.

I have said since $a_n\rightarrow a$ there exists a $N$ such that for every $n>N$ we have $|a_n-a|<\epsilon_0$. Therefore we have for $n>N$

$|\frac{1}{n}\cdot(a_1+...+a_N+...a_n)-a|=|\frac{1}{n}\cdot(a_1+...+a_N+...a_n)-\frac{na}{n}|=|\frac{a_1-a}{n}+...+\frac{a_{N+1}-a}{n}+...+\frac{a_n-a}{n}|\Longrightarrow -\frac{n\epsilon_0}{n}-|\frac{a_1-a}{n}+...+\frac{a_N-a}{n}|<|\frac{1}{n}\cdot(a_1+...+a_N+...a_n)-\frac{na}{n}|<\frac{n\epsilon_0}{n}+|\frac{a_1-a}{n}+...+\frac{a_N-a}{n}|$

We choose now $n$ so big such that for $n>N'$ $|\frac{a_1-a}{n}+...+\frac{a_N-a}{n}|<\epsilon_1$

Because $\epsilon_1$ and $\epsilon_0$ were arbitrary the claim is proved.

If there are two points for this excercise how many would you give me?


marked as duplicate by Nosrati, Mark Viola real-analysis Apr 15 at 16:55

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  • $\begingroup$ I think it could look a little cleaner but the idea is definitely right. $\endgroup$ – Clayton Apr 15 at 16:27
  • $\begingroup$ The problem with the proof is that I have said $\frac{n\epsilon}{n}$ altough I should have said $\frac{(n-N)\epsilon}{n}$. The proof is still corect if I choose an appropriate $N''$ but the prof might give me no points for this blunder $\endgroup$ – New2Math Apr 15 at 16:35
  • 2
    $\begingroup$ $(n-N) \leq n$, so your estimation is correct. I want to point out that you don't need a lower bound since the absolute value is greater than or equal to $0$ anyway. However, I would mark you full points. $\endgroup$ – Nathanael Skrepek Apr 15 at 16:40