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Where $a_n \in C$, intuitively I think this must be true as if are taking an average of almost all arbitrarily small numbers then the average will be arbitrarily small.

let $\epsilon>0, \exists N:|a_n|<\epsilon, \forall n>N$ $$ |(a_1+\cdots+a_n)/n|<\epsilon$$ $$|a_1+\cdots+a_N+\cdots+a_n|<e n$$ I know this is wrong but this was what I was thinking $$|a_1+\cdots+a_N|+(n-N)\epsilon<en$$ Could I have a rough hint on a how to better estimate the value of $ |(a_1+\cdots+a_n)/n|$

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marked as duplicate by user99914, clark, Guy Fsone, ahulpke, Namaste Jan 27 '18 at 0:28

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  • $\begingroup$ No the way you start looks exactly right... not sure why you put the $n$ on the RHS. The first term tends to zero and the second term is less than epsilon so it can always be less than epsilon for sufficiently large $n.$ $\endgroup$ – spaceisdarkgreen Jan 26 '18 at 23:12
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Let $\varepsilon>0$. There exists an integer $N_0$ such that $$|a_k|<\frac\varepsilon 2 \quad\forall k>N_0.$$ So, if $n>N_0$, we have \begin{align} \biggl|\frac{a_1+\dots+a_{N_0}+\dots +a_n}n\biggr|&\le\frac{|a_1+\dots+a_{N_0}|}n+\frac{|a_{N_0+1}|}n+\dots+\frac{|a_{n}|}n \\[0.5ex] &< \frac{|a_1+\dots+a_{N_0}|}n+\frac{n-N_0}n\frac\varepsilon 2 <\frac{|a_1+\dots+a_{N_0}|}n +\frac\varepsilon 2 \end{align} Now $|a_1+\dots+a_{N_0}|$ does not depend on $n$, so it tends to $0$ as $n$ tends to $\infty$, and there exist an integer $N_1$ such that $$\frac{|a_1+\dots+a_{N_0}|}n<\frac\varepsilon 2 \quad\forall n>N_1. $$ Now set $N=\max(N_0,N_1)$. For all $n>N$, both inequalities are satisfied, so $$\biggl|\frac{a_1+\dots+a_N+\dots a_n}n\biggr|<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon.$$

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