5

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:

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

Pedro
  • 122,002

2 Answers2

4

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

AD - Stop Putin -
  • 10,970
  • Ok, but use your $n_0$ then, it results in the same. – AD - Stop Putin - Oct 10 '12 at 21:00
  • Your estimate of the tail is the problem, because the tail is positive and the upper bound of yours is negative. You should simply estimate the terms by $\varepsilon$ and count them. – AD - Stop Putin - Oct 10 '12 at 21:04
  • 1
    I'm just getting $$\frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} < \frac{{n - N}}{n}\frac{\epsilon }{2}$$ Why is that wrong? I think your $n-N-1$ should be a $n-(N+1)+1=n-N$ – Pedro Oct 10 '12 at 21:07
  • OK, I can go with $\limsup$s now. – Pedro Oct 11 '12 at 02:11
  • Right, yes I had an extra term which I just removed. The estimate of the comment is correct, but in the OP we have $$\frac{1}{n}\sum\limits_{k = N + 1}^n {\left| {{a_k} - \ell } \right|} \leq - \frac{{N }}{2n}\epsilon-\epsilon/2 $$ which is certainly not the case. – AD - Stop Putin - Oct 11 '12 at 05:40
  • Oh, the last term should have a plus sign. See the second time I write that out. – Pedro Oct 11 '12 at 09:39
  • @PeterTamaroff I see! – AD - Stop Putin - Oct 11 '12 at 16:27
1

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 $$\hat a_n\to\ell.$$

leo
  • 10,433
  • Just rewriting your argument What for? – Did Apr 28 '13 at 06:39
  • @Did to provide some justification on some of the steps of the OP original argument. Basically reordering ideas... – leo Apr 28 '13 at 06:50
  • Really? Then you might want to make explicit which steps need more justification. // As an aside, the fact that your $M$ exists is not primarily a consequence of the convergence you cite. And using such a bound valid for every $n$ seems to only complicate things. – Did Apr 28 '13 at 06:54
  • @Did the sequence at the beginning is convergent, so it's bounded. The $M$ is just one such bound. Don't see how its validity for all $n$ can complicate things. Is there something wrong in the above? – leo Apr 28 '13 at 07:14
  • (1.) I do not know what "the sequence at the beginning is convergent" means. (2.) You did not explain what the problem with the other answer is. (3.) I did not use the word "wrong". – Did Apr 28 '13 at 07:29
  • There is no problem with the other answer. I'll try to improve this in a few hours. I'll go to sleep – leo Apr 28 '13 at 07:38
  • @Did OK. Nevermind. – Pedro Apr 28 '13 at 13:56