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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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    Oh, I see, I feel dumb right now, haha Thanks!2012-08-05

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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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    I am happy to help. By the way, I wrote $|x_n|^2$ since the above works for complex $x_n$ too.2012-08-07
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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.