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I am trying to show that for each sequence of real numbers $\{a_k\}_{k = 1}^\infty$, with $\sum_{k=1}^\infty |a_k| < \infty$, we also have that $\sum_{k=1}^\infty |a_k|^2 < \infty$.

In my "proof", so far I have deduced that for each positive integer $n$, we have \begin{align} \sum_{k=1}^n |a_k|^2 \leq \left(\sum_{k=1}^\infty |a_k|\right) \left( \sum_{k=1}^n |a_k|\right). \end{align} I am tempted to say that taking the limit as $n \to \infty$, we have \begin{align} \sum_{k=1}^\infty |a_k|^2 \leq \left(\sum_{k=1}^\infty |a_k|\right)^2, \end{align} but I am not completely convinced of this step. The reason is that all we seem to know is that the partial sums $\sum_{k=1}^n |a_k|^2$ are bounded, but this this alone is not sufficient to deduce convergence.

Is it possible to extend the inequality regarding the partial sums to deduce the convergence of $\sum_{k=1}^\infty |a_k|^2$?

  • $a_k \to 0$, so that $|a_k|^2 \le |a_k|$ for all but finitely many $k$. – This must have been asked and answered before. – Martin R Jul 04 '19 at 12:05
  • @MartinR Whoops, looks like this has been asked before more than once. I honestly tried to find a similar question, but couldn't. What should I do? –  Jul 04 '19 at 12:08
  • There should be a button where you can confirm that the question is a duplicate. Or just wait – I assume that it will be closed by other people with sufficient reputation. – Martin R Jul 04 '19 at 12:11
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    What? Of course $s_n=\sum_1^n|a_j|^2$ bounded implies convergence. The partial sums are a non-decreasing sequence, bounded above, hence convergent. – David C. Ullrich Jul 04 '19 at 15:14
  • @DavidC.Ullrich thanks for that, I didn't know of that result. I'm glad you made me aware of it. –  Jul 05 '19 at 03:35

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