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I am having this:

$ (\ x_n)\ _{n \in \mathbb N} $ is sequence in $\mathbb C$, so the series $\sum_{n=1}^{\infty} |x_n|$ converges.

I've already proved that the series $\sum_{n=1}^{\infty} |x_n|^2$ converges with the cauchy-product:

$\sum_{n=1}^{\infty} |x_n| * \sum_{n=1}^{\infty} |x_n| = \sum_{n=1}^{\infty}\sum_{k=0}^{n}|x_k|*|x_{n-k}|\ge \sum_{n=1}^{\infty}\sum_{k=0}^{n}|x_k|*|x_{n-k}|=\sum_{n=1}^{\infty} |x_n|^2$


Now I have to prove that:

$( \sum_{n=1}^{\infty} |x_n|^2)^{1/2} \le \sum_{n=1}^{\infty} |x_n|$


Questions:

  1. Is my Proof correct?
  2. How can I prove the other question?
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    $\begingroup$ Apply Cauchy-Schwartz inequality for $a_n=\frac{x_n}{n}$ and $b_n=1$. $\endgroup$ – Volodymyr Fomenko Jan 11 '15 at 19:56
  • $\begingroup$ @VolodymyrFomenko We do not use the Cauchy-Schwartz inequality. $\endgroup$ – basti12354 Jan 11 '15 at 20:00
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Let $S^{(1)}_m = \sum_{k=1}^m|x_k|$ and $S^{(2)}_m = \sum_{k=1}^m|x_k|^2$. Assume for induction that the inequality $\sqrt{S_m^{(2)}} \leq S_m^{(1)}$ holds for $m=1,2,\ldots,n$. Then

$$\sqrt{S_{n+1}^{(2)}} = \sqrt{S_n^{(2)}+|x_{n+1}|^2} \leq \sqrt{(S_n^{(1)})^2 + |x_{n+1}|^2} \leq S_n^{(1)} + |x_{n+1}| = S_{n+1}^{(1)}$$

so the inequality holds for $m=n+1$. Since $\sqrt{|x_1|^2} \leq |x_1|$ we get that the inequality holds for all $m\in\mathbb{N}$.


A simpler and more direct approach is to just compute

$$\left(\sum_{k=1}^n|x_k|\right)^2 = \sum_{k=1}^n|x_k|^2 + \sum_{i\not= j}|x_ix_j| \geq \sum_{k=1}^n|x_k|^2$$

since $|x_ix_j| \geq 0$.

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Hint: Prove that $$\left(\sum_{n=1}^{N} |x_n|\right)^2 \ge\sum_{n=1}^{N} |x_n|^2$$ for all $N\in\mathbb{N}$.

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  • $\begingroup$ $(\sum_{n=1}^{\infty} |x_n|)^2=\sum_{n=1}^{\infty} |x_n| * \sum_{n=1}^{\infty} |x_n| = \sum_{n=1}^{\infty}\sum_{k=0}^{n}|x_k|*|x_{n-k}|\ge \sum_{n=1}^{\infty}\sum_{k=0}^{n}|x_k|*|x_{n-k}|=\sum_{n=1}^{\infty} |x_n|^2$ $\endgroup$ – basti12354 Jan 11 '15 at 20:15
  • $\begingroup$ But how can I use this for my second proove? $\endgroup$ – basti12354 Jan 11 '15 at 20:16
  • $\begingroup$ Better use induction to prove this. $\endgroup$ – Volodymyr Fomenko Jan 11 '15 at 20:18
  • $\begingroup$ @VolodymyrFomenko Should I use Induction to proove the answear of "Sameer" or to proove my question? $\endgroup$ – basti12354 Jan 11 '15 at 20:24

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