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If I let $l_2$ be the set of all real sequences $\{x_j\}_{j\in N}$, such that

$\sum_{j=1}^{\infty} x_j^2 < \infty$,

is there any way to show that this sum converges? Can I do it by showing that $\{x_j\}_{j\in N}$ is Cauchy?

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  • $\begingroup$ What sum?${}{}{}$ $\endgroup$ – Saucy O'Path Apr 9 at 13:12
  • $\begingroup$ $\sum_{j=1}^{\infty} x_j^2$ $\endgroup$ – Samsam22 Apr 9 at 13:19
  • $\begingroup$ So the question is: "Does $\sum_{j=1}^\infty$ converge if $x\in S$, provided that $S$ is the set of sequences such that $y\in S$ if and only if $\sum_{j=1}^\infty y_j^2$ converges to a real number?" $\endgroup$ – Saucy O'Path Apr 9 at 14:14
  • $\begingroup$ What I'm really supposed to find out is that if $\{x_j\},\{y_j\}\in l_2$, then $\sum_{j=1}^{\infty} x_j y_j$ converges. I thought the best way of showing this is to first show that $\sum_{j=1}^{\infty}x_j^2$ converges, and then use Cauchy-Schwartz inequality $\endgroup$ – Samsam22 Apr 10 at 13:34
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Let $S_k = \sum_{j=1}^k x_j^2$. By definition the sum $\sum_{j=1}^\infty x_j^2$ converges if and only if $S_k$ converges.

Since $x_i^2>0$, the sequence $S_k$ is monotonically increasing, by assumption it is also bounded and so it converges. In fact, $S_k\rightarrow S$ where $S=\sup \{S_k : k\in\mathbb{N}\}$.

Note that this implies that $x_j^2\rightarrow 0$, hence $x_j\rightarrow 0$. However the fact that $x_j\rightarrow 0$ does not imply that $S_k$ converges.

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