If $a_n\to \ell$ then $\hat a_n\to \ell$ [duplicate]

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:

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

marked as duplicate by Guy Fsone, user99914, José Carlos Santos, Sahiba Arora, Did real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 1 '18 at 22:43

• Why $\left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| = \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - \ell } \right)} } \right|$ instead this: $\left| {\frac{1}{n}\sum\limits_{k = 1}^n {{a_k}} - \ell } \right| = \left| {\frac{1}{n}\sum\limits_{k = 1}^n {\left( {{a_k} - n \ell } \right)} } \right|$? – M. Strochyk Oct 10 '12 at 20:57
• @M.Strochyk That is wrong. The $1/n$ kills it off. – Pedro Tamaroff Oct 10 '12 at 20:58
• @M.Strochyk Because that gives too many $\ell$'s – AD. Oct 10 '12 at 20:59
• This question is about the same result: Prove convergence of a sequence. Perhaps there and in linked questions several proofs (or several versions of the same proof) can be found. AFAIK the proofs I've seen are not that different from the proof of Stolz-Cesaro. The question limit of quotient of two series has relatively detailed proof of Stolz-Cesaro, your question Stolz-Cesàro Theorem contains several links. – Martin Sleziak Oct 11 '12 at 6:23
• @MartinSleziak Yes, I believe it is a popular question, here is one more Arithmetic mean sequence of a convergent complex sequence. – AD. Oct 11 '12 at 9:39

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

• Ok, but use your $n_0$ then, it results in the same. – AD. 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. Oct 10 '12 at 21:04
• 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 Tamaroff Oct 10 '12 at 21:07
• OK, I can go with $\limsup$s now. – Pedro Tamaroff Oct 11 '12 at 2:11
• 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. Oct 11 '12 at 5:40

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

• Just rewriting your argument What for? – Did Apr 28 '13 at 6:39
• @Did to provide some justification on some of the steps of the OP original argument. Basically reordering ideas... – leo Apr 28 '13 at 6:50
• 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 6:54
• @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 7:14
• (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 7:29