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I have been trying to prove this for the last three days. It's for my Time Series Econometrics homework. I think you'll notice I'm not good at math as I can't even express my solution very well, and I'm sorry for that.

At first I thought that if: (1) every element is finite; (2) there is one element that is greater or equal to any other in the sequence; (3) no element repeats itself for more than a finite number of times, then we would have $\sum \limits_{i=0}^{\infty}{|x_i|}<\infty$, because the terms would converge to zero(sooner or later, hehe).

Believing that $x_i^2$ is a monotonic transformation over $|x_i|$, all the properties that guaranteed the convergence of a series would remain unaltered and thus the proof would be given that $\sum \limits_{i=0}^{\infty}{x_i^2}<\infty$.

But now I've just seen that the convergence of terms to zero does not guarantee the convergence of a series, and got pretty confused about the correct approach to the problem.

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Hint: if $\sum |x_i| < \infty$ then $|x_i|$ approaches $0$. Thus only a finite number of terms are $\geq 1$. What happens to a small number when you square it? – Karolis Juodelė Aug 5 '12 at 18:22
Oh, I see, I feel dumb right now, haha Thanks! – Fbrufino Aug 5 '12 at 18:30
up vote 2 down vote accepted

Here is an alternative proof.

First note that since $\sum |x_n|$ converges the sequence $\{|x_n|\}$ is bounded. The least upper bound $\{|x_n|\}$ is usually denoted by $\sup_k |x_k|$.

Secondly we use $|x_n|^2\leq \sup_k|x_k|\cdot|x_n|$ to reach $$\left|\sum x_n^2\right|\leq\sum |x_n|^2\leq \sup|x_k|\cdot\sum |x_n|$$ which does not only imply convergence, but also a general estimate.

Edit: Perhaps the convergence is a bit hidden: The key is that above estimate shows that the tail of the series tends to 0, that is $$ \left|\sum_{n=N}^\infty x_n^2\right|\leq\sum_{n=N}^\infty |x_n|^2\leq \sup|x_k|\cdot\sum_{n=N}^\infty |x_n| \to0\qquad \text{as $N\to\infty$} $$

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Thanks, I actually needed some help with the formalization too – Fbrufino Aug 5 '12 at 21:46
I am happy to help. By the way, I wrote $|x_n|^2$ since the above works for complex $x_n$ too. – AD. Aug 7 '12 at 19:56

Since $\sum\limits_n|x_n|$ converges, $x_n\to0$ hence there exists a finite $N$ such that $|x_n|\leqslant1$ for every $n\geqslant N$. Thus, $\sum\limits_nx_n^2\leqslant\sum\limits_{n\lt N}x_n^2+\sum\limits_{n\geqslant N}|x_n|$ is finite.

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