# Show that if $a_n\rightarrow a$ then $\frac{a_1+…+a_n}{n}\rightarrow a$ [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 15 at 16:55

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• I think it could look a little cleaner but the idea is definitely right. – Clayton Apr 15 at 16:27
• 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 – New2Math Apr 15 at 16:35
• $(n-N) \leq n$, 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. – Nathanael Skrepek Apr 15 at 16:40