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Good night. I'm stuck in this prove.

Prove if $\left\{ a_{n}\right\} $ is a sequence and $$\lim_{n\rightarrow\infty}a_{n}=L$$ then $$\lim_{n\rightarrow\infty}\frac{a_{1}+a_{2}+...+a_{n}}{n}=L$$

I make this:

$\mid\frac{a_{1}+...+a_{n}}{n}-L\mid=\mid\frac{a_{1}+...+a_{n}-nL}{n}\mid\leq\frac{\mid a_{1}-L\mid+...+\mid a_{N}-L\mid+\mid a_{N+1}-L\mid+...+\mid a_{n}-L\mid}{n}=\frac{\mid a_{1}-L\mid+...+\mid a_{N-1}-L\mid}{n}+\frac{\mid a_{N}-L\mid+\mid a_{N+1}-L\mid+...+\mid a_{n}-L\mid}{n}$

And I'm stuck, please help!!

Martin Sleziak
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rcoder
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  • Use the fact that for any $\epsilon>0$ there exists an $N>0$ such that $|a_n-L|<\epsilon$ for all $n>N$. – Alex R. Aug 02 '16 at 22:41
  • Also here. You may find even more on the "related" column. –  Aug 02 '16 at 22:44
  • One of those fractions has a finite number of terms in the numerator, and hence is bounded. As $n$ gets large it is less than $\frac \epsilon 2$ the other fraction, all of the terms in the numerator are less than $\frac \epsilon 2$, and there are fewer than $n$ of them. – Doug M Aug 02 '16 at 23:11
  • Can't you partition the tail of your se uence into monotone se uences? Their running averages will each have the same monoticity. Thus they each converge and the particular limit is obvious. – Jacob Wakem Aug 03 '16 at 00:48
  • Intuitively, the running-average is almost-strictly a dampening of your sequence. Thus its convergence behavior is almost-strictly better. the particular limit is obvious. – Jacob Wakem Aug 03 '16 at 01:17

1 Answers1

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You are almost there.

In your split sum, $\mid\frac{a_{1}+...+a_{n}}{n}-L\mid=\mid\frac{a_{1}+...+a_{n}-nL}{n}\mid\leq\frac{\mid a_{1}-L\mid+...+\mid a_{N}-L\mid+\mid a_{N+1}-L\mid+...+\mid a_{n}-L\mid}{n}=\frac{\mid a_{1}-L\mid+...+\mid a_{N-1}-L\mid}{n}+\frac{\mid a_{N}-L\mid+\mid a_{N+1}-L\mid+...+\mid a_{n}-L\mid}{n} $,

choose $N$ large enough as a function of $\epsilon$ so that $\mid a_{k}-L\mid \lt \epsilon$ for $k > N$ and then choose $n$ large compared to $N$ so that $\frac{\mid a_{1}-L\mid+...+\mid a_{N-1}-L\mid}{n} \lt \epsilon $.

The resulting sum is less than $2\epsilon$.

marty cohen
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