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I'm wondering if I have a sufficient proof of the following:

If $(a_n)$ is a sequence such that $\lim_{n \rightarrow \infty}a_n=A$, then $\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A$.

My approach:

For all $\varepsilon > 0$, there exists $N$ such that for all $k>N$, $|a_k -A|<\varepsilon$. So we can break the limit up as follows $$\lim_{n\rightarrow \infty} \frac{a_1+...+a_k}{n} + \lim_{n \rightarrow \infty}\frac{a_{k+1}+...+a_n}{n} \overset{\epsilon \rightarrow 0}{=} 0 + \lim_{n \rightarrow \infty}\frac{nA}{n}=A$$

Is this on the right track, or am I missing something about breaking up the limit in the way I have?

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marked as duplicate by Martin Sleziak, Cameron Williams, Daniel Fischer real-analysis Oct 16 '16 at 9:50

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  • $\begingroup$ Have you tried using the squeeze theorem between: $\frac{n}{\sum_{r=1}^{n}\frac{1}{a_r}}>\frac{a_1+a_2+...+a_n}{n}>(a_1a_2...a_n)^\frac{1}{n}$ $\endgroup$ – J.Gudal Dec 10 '15 at 3:34
  • $\begingroup$ This is the intuition for the solution, but it should probably be made more rigorous with some (simple) epsilon pushing. You have the standard trick of splitting a quantity into two parts you know how to control, now just make each less than $\varepsilon/2$. $\endgroup$ – user217285 Dec 10 '15 at 3:35
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  • $\begingroup$ Your assume that $\lim_{n\to \infty}(a_{k+1}+...+a_n)/n$ exists and is equal to $A$ which is circular reasoning. $\endgroup$ – DanielWainfleet Oct 16 '16 at 8:40
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Since $(a_n)$ is convergent hence it's bounded. Let $|a_n|\leq K\ \forall\ n\in \mathbb{N}$. Now for $\epsilon >0$ let $N\in \mathbb{N}$ be such that $|a_n-A|<\epsilon\ ,\forall\ n>N$.

Consider $|\frac{a_1+a_2+...+a_n}n-A|=|\frac{(a_1-A)+(a_2-A)+...+(a_n-A)}n|\leq\frac{|a_1-A|}n+\frac{|a_2-A|}n+...+\frac{|a_n-A|}n$.

Now choose $M\in \mathbb{N}$ (What $M$ ?) and bound the first $M$ terms of the above expression using the boundedness of $(a_n)$ and the rest of the terms using the fact that $|a_n-A|\rightarrow0$ as $n\rightarrow \infty$.

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  • $\begingroup$ Do we take M=N? That way, we get the required as $\le K\frac{N}{n} + \varepsilon\frac{n-N}{n}$ I think I am v wrong, please explain? $\endgroup$ – Za Ira Dec 16 '18 at 2:55
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For any given $\epsilon>0$, choose $N$ such that $A-\epsilon/2<a_k<A+\epsilon/2$ for all values of $k>N$. Then, whenever $n>\max\left(N,\frac{2\left|\sum_{k=1}^N(a_k-A)\right|}{\epsilon}\right)$ we have

We have

$$\begin{align} \left|\frac1n\sum_{k=1}^na_k-A\right|&=\left|\frac1n\sum_{k=1}^n(a_k-A)\right|\\\\ &\le \left|\frac1n\sum_{k=1}^N(a_k-A)\right|+\left|\frac1n\sum_{k=N+1}^n(a_k-A)\right|\\\\ &\le \left|\frac1n\sum_{k=1}^N(a_k-A)\right|+\frac1n\sum_{k=N+1}^n\left|a_k-A\right|\\\\ &\le \frac{\epsilon}{2}+\frac{\epsilon}{2}\left(1-\frac Nn\right)\\\\ &< \epsilon \end{align}$$

And we are done!

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. - Mark $\endgroup$ – Mark Viola Dec 16 '15 at 19:37
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Given $e>0,$ choose the least (or any) $k$ such that $i>k \implies |a_i-A|<\frac {e}{3}.$

Now choose $m$ large enough that $m>k$ and $\frac {k}{m}|A|<\frac {e}{3}$ and $m^{-1}|\sum_{i=1}^ka_i|<\frac {e}{3} .$

For $n\geq m$ we have $$|-nA+\sum_{i=1}^na_i|= |\sum_{i=1}^ka_i+kA+\sum_{i=k+1}^n(a_i-A)|<$$ $$<\frac {em}{3}+k|A|+\sum_{i=k+1}^n|a_i-A|<\frac {em}{3}+k|A|+(n-k)\frac {e}{3}.$$ So $n\geq m$ implies $$|-A+n^{-1}\sum_{i=1}^na_i|\; <\; \frac {e}{3} \frac {m}{n}+\frac {m}{n}\cdot \frac {k}{m}|A|+(1-k/n)\frac {e}{3}<e.$$

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  • $\begingroup$ To the OP; You are on the right track. One more step was needed: Re-arrange $-A+n^{-1}\sum_{i=k+1}^na_i$ as $[n^{-1}\sum_{i=k+1}^n(a_i-A)] + kA/n.$ $\endgroup$ – DanielWainfleet Oct 16 '16 at 9:37
  • $\begingroup$ A slightly different style would be to show that $-e< \lim_{m\to \infty}\inf_{n>m}f(n) \leq$ $ \lim_{m\to \infty}\sup_{n>m}f(n)<e$ for every $e>0,$ where $f(n)=-A+n^{-1}\sum_{i=1}^na_i.$ $\endgroup$ – DanielWainfleet Oct 16 '16 at 9:45

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