$\lim\limits_{i \to \infty}a_i=a$ prove $\lim\limits_{N\to \infty} \sum\limits_{i=1}^N \frac{a_i}{N}=a$

Question

So The question is like in the title I tried some algebra of limits, but Frankly speaking I really have no clue where to start especially, because if $a\neq 0 $ then $\sum\limits_{i=1}^{\infty} {a_i} = a$ diverges. So just a hint where to start, what to look at would be great.

score 1 · Accepted Answer · answered Sep 14 '18 at 09:21

Hint: $\frac 1 N \sum_{i=1}^{N} a_i -a =\frac 1 N \sum_{i=1}^{N} (a_i-a)$ $=\frac 1 N \sum_{i=1}^{k} (a_i-a)+\frac 1 N \sum_{i=k+1}^{N} (a_i-a)$. Make the second term small for a suitable $k$ using the fact that $a_n \to a$ and then take limit in the first term.

$\lim\limits_{i \to \infty}a_i=a$ prove $\lim\limits_{N\to \infty} \sum\limits_{i=1}^N \frac{a_i}{N}=a$

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