# Convergence of $\sum_{j=1}^\infty x_j^2$ assuming that the sum of squares is finite.

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?

• What sum?${}{}{}$ – Saucy O'Path Apr 9 at 13:12
• $\sum_{j=1}^{\infty} x_j^2$ – Samsam22 Apr 9 at 13:19
• 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?" – Saucy O'Path Apr 9 at 14:14
• 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 – Samsam22 Apr 10 at 13:34

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.