1
$\begingroup$

Let $\{a_i\}_{i\in\mathbb{Z}}$ be some real sequence and let $S_n = \frac{1}{n}\sum_{|i|<n}^n|a_i|$. I need to verify two claims.

First, if $a_n\to0$ as $n\to\infty$, why $$\lim_{n\to\infty}\frac{1}{n}S_n=2\lim_{n\to\infty}|a_i|=0$$

Second, it is claimed that by the dominated convergence theorem $$\lim_{n\to\infty}\sum_{|i|<n}\left(1 - \frac{|i|}{n}\right)a_i = \sum_{i\in\mathbb{Z}}a_i$$

Could you help me, please to see why this is the case, why do we need the dominated convergence theorem and what is the dominating sequence.

$\endgroup$
1
$\begingroup$

For the first part, assuming $a_n\to 0$ as $|n|\to \infty$, note that $$\frac{S_n}{n}\le \frac{2\max_{i\in \{-n,\cdots,n\}}|a_i|}{n}\le \frac{2M}{n}$$ where $M$ is upper bound on $\{|a_n|\}$, which exists since $\{a_n\}$ converges. So, $\lim_{n\to \infty}S_n/n=2\lim_ {n\to \infty}|a_i|=0$.

$\endgroup$
  • $\begingroup$ Why should $\lim_{n \to \infty} a_n = 0$ imply that $\sup_{i \in \mathbb{Z}} |a_i| < \infty$? $\endgroup$ – GenericNickname Apr 22 '15 at 7:49
  • $\begingroup$ @GenericNickname, this is because, as $a_n\to 0$, $\forall \epsilon>0,\ \exists N$ such that $|a_n|<\epsilon\ \forall n\ge N$. $\endgroup$ – Samrat Mukhopadhyay Apr 22 '15 at 7:54
  • $\begingroup$ Yes. But it does not prevent $a_n \to \infty$ for $n \to -\infty$, or am I missing something? $\endgroup$ – GenericNickname Apr 22 '15 at 7:55
  • $\begingroup$ Yes, @GenericNickname, you are correct. I think for the result in the question to hold true, we need $a_n\to 0$ as $|n|\to \infty$. Otherwise, we could have $\sup_{i\in \{-n,\cdots,n\}}|a_n|>\epsilon$, for some $\epsilon$ for infinitely many $n's$. $\endgroup$ – Samrat Mukhopadhyay Apr 22 '15 at 7:58

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.