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I need some help to finish this proof:

THEOREM

Let $\{a_n\}$ be such that $\lim a_n=\ell$ and set

$\hat a_n=\frac 1 n \sum_{k=1}^na_k$

Then $\lim\hat a_n=\ell$

PROOF

Let $\epsilon >0 $ be given. Since $\lim a_n=\ell$ , there exists an $N$ for which $\left| {{a_n} - \ell } \right| < {\epsilon/2 }$

whenever $n>N$. Now:

$\begin{eqnarray*} \left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| &=& \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - \ell } \right)} } \right| \\ &=& \left| {\frac{1}{n}\sum\limits_{k = 1}^N {\left( {{a_k} - \ell } \right)} + \frac{1}{n}\sum\limits_{k = N + 1}^n {\left( {{a_k} - \ell } \right)} } \right| \\ &\leqslant& \frac{1}{n}\sum\limits_{k = 1}^N {\left| {{a_k} - \ell } \right|} + \frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} \\ & <& \frac{N}{n}\zeta - \frac{{N }}{2n}\epsilon+\epsilon/2 \end{eqnarray*} $

where $\zeta=\mathop {\max }\limits_{1 \leqslant k \leqslant N} \left| {{a_k} - \ell } \right|$

Now, let $n_0$ be such that if $n>n_0$,

$\eqalign{ & \frac{{N\zeta }}{n} < {\epsilon} \cr & \frac{N}{n} < {1} \cr} $ Then we get

$\frac{N}{n}\zeta - \frac{N}{n}\frac{\epsilon }{2} + \frac{\epsilon }{2} < \epsilon - \frac{\epsilon }{2} + \frac{\epsilon }{2} = \epsilon $

How far is this OK? Do you think there is an easier way to go about proving it? I now remember that by Stolz Cesàro:

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\sum\limits_{k = 1}^{n + 1} {{a_k}} - \sum\limits_{k = 1}^n {{a_k}} }}{{n + 1 - n}} = \mathop {\lim }\limits_{n \to \infty } {a_{n + 1}} = \ell $

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    @MartinSleziak Yes, I believe it is a popular question, here is one more [Arithmetic mean sequence of a convergent complex sequence](http://math.stackexchange.com/questions/207690/arithmetic-mean-sequence-of-a-convergent-complex-sequence).2012-10-11

2 Answers 2

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For the tail sum $\frac1n\sum_{N+1}^n|a_k-\ell|\leq \frac{(n-N)}{n}\varepsilon<\varepsilon$ for all large $n$, hence $\limsup_n \frac1n\sum_{1}^n|a_k-\ell|\leq 0+\varepsilon=\varepsilon...$

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    @PeterTamaroff I see!2012-10-11
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Just rewriting your argument:

The sequence $\{|a_n-\ell|\}$ converges to $0$ therefore it's bounded, so there exist some $M\gt 0$ such that $|a_n-\ell|\leq M\quad\forall n\in \Bbb N.$ This $M$ plays the role of the $\zeta$ in your proof, but notice it does not depends on $n$.

Let $\epsilon\gt 0$. There exist $n_1\in\Bbb N$ such that for all $n\in\Bbb N$, $n\geq n_1$ implies $|a_n-\ell|\lt\frac{\epsilon}{2}.$

Since $\lim_{n\to\infty} \frac{n_1}{n}=0,$ there exist $n_2\in\Bbb N$ such that, for all $n\in\Bbb N$, $n\geq n_2$ implies $\frac{n_1}{n}\leq \frac{\epsilon}{2M}.$

Let $N=\max\{n_1,n_2\}$, then for all $n\in\Bbb N$, $n\geq N\geq n_1$ implies $\left(1-\frac{n_1}{n}\right)\leq 1$ and then $\begin{align*} \left| -\ell+\frac1{n}\sum_{j=1}^n a_j \right| &\leq \frac1{n}\sum_{j=1}^n \left|a_j-\ell\right|\\ &= \frac1{n}\sum_{j=1}^{n_1} \left|a_j-\ell\right| + \frac1{n}\sum_{j=n_1+1}^{n} \left|a_j-\ell\right|\\ &\lt \frac1{n} n_1M + \frac1{n}(n-n_1)\frac{\epsilon}{2}\\ &=\frac{n_1}{n}M + \left(1-\frac{n_1}{n}\right)\frac{\epsilon}{2}\\ &\lt \frac{\epsilon}{2M}M + 1\cdot \frac{\epsilon}{2}=\epsilon. \end{align*}$

Therefore $\hat a_n\to\ell.$

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    @Did OK. Nevermind.2013-04-28