1
$\begingroup$

Let $(x_j)\to x$ a sequence on $\Bbb K$, then show that $(\frac1n\sum_{j=1}^n x_j)\to x$

I want to check if this proof for the convergence of Césaro sums is valid. I rewrite the statement to prove as $(\frac1n\sum_{j=1}^n (x_j-x))\to 0$.

Because we have that $(x_j-x)\to 0$ then for any $\epsilon>0$ exists some $N\in\Bbb N$ such that $|x_j-x|<\epsilon$ when $j\ge N$.

Then we have that for $n\ge N$ then

$$\left|\frac1n\sum_{j=1}^{n}(x_j-x)\right|\le\frac1n\sum_{j=1}^{n}|x_j-x|<\frac1n\sum_{j=1}^{N-1}|x_j-x|+\frac1n\sum_{j=N}^{n}\epsilon$$

Then

$$\lim_{n\to\infty}\left|\frac1n\sum_{j=1}^{n}(x_j-x)\right|\le\lim_{n\to\infty} \left(\frac1n\sum_{j=1}^{N-1}|x_j-x|+\frac1n\sum_{j=N}^{n}\epsilon\right)=\epsilon$$

Because $\epsilon$ can be chosen arbitrarily small we conclude that $(\frac1n\sum_{j=1}^n (x_j-x))\to 0$, and hence

$$(x_j)\to x\implies \left(\frac1n\sum_{j=1}^n x_j\right)\to x$$


What I dont like of this proof, if correct, is that Im unable to show the convergence of the Cesáro sum with the classic language of $\epsilon-N$, what I did above is a different thing. This is the reason why Im unsure about the correctness of this proof.

Link to similar question: here

$\endgroup$
  • 1
    $\begingroup$ Who is $\mathbb K$? $\endgroup$ – Math1000 Oct 14 '16 at 22:55
  • $\begingroup$ $\Bbb K$ is the common notation for $\Bbb R$ or $\Bbb C$ as the set where the $x_j$ are. It is useful to say that the statement is true for reals or complex numbers. $\endgroup$ – Masacroso Oct 14 '16 at 22:58
  • 2
    $\begingroup$ The proof is correct. $\endgroup$ – Paolo Leonetti Oct 14 '16 at 23:03
  • $\begingroup$ @JohnMa Im asking for a confirmation for the way I did this proof, so it cannot be a duplicate. $\endgroup$ – Masacroso Oct 14 '16 at 23:44
  • $\begingroup$ Your proof is no difference than most of the others, and indeed you accepted an answer which is essentially the answer given in the duplicate. BTW, in your answer you should write $\limsup$ instead of $\lim$ since you do not know yet a limit exists. @Masacroso $\endgroup$ – user99914 Oct 15 '16 at 0:02
1
$\begingroup$

With some minor modification you have the $\epsilon - N$ proof.

If $n > N$ we have $|x_n -x| < \epsilon/2$.

Hence,

$$\left|\frac{1}{n}\sum_{j=1}^n x_j - x \right| \leqslant \frac{1}{n} \sum_{j=1}^N|x_j -x| + \frac{1}{n} \sum_{j= N+1}^n |x_j - x| \leqslant \frac{1}{n} \sum_{j=1}^N|x_j -x| + (1 - N/n)\frac{\epsilon}{2}$$

Now with $N$ fixed, choose $n$ sufficiently large such that

$$\frac{1}{n} \sum_{j=1}^N|x_j -x|< \frac{\epsilon}{2}.$$

Since $1 - N/n < 1$, we have for $n$ sufficiently large

$$\left|\frac{1}{n}\sum_{j=1}^n x_j - x \right| < \epsilon$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.