Suppose that $\displaystyle\sum_{n=1}^{\infty}\ a_n$ is absolutely convergent. How can we prove that $\displaystyle\sum_{n=1}^{\infty}\ a_n^2$ is convergent?
Prove that $\sum_{n=1}^{\infty}\ a_n^2$ is convergent if $\sum_{n=1}^{\infty}\ a_n$ is absolutely convergent
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real-analysis
sequences-and-series
2 Answers
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$\lim_{n \to \infty} \frac{a^2_n}{|a_n|} = \lim_{n \to \infty} |a_n| = 0,$ since $\sum |a_n|$ converges. By the limit comparison test, $\sum a^2_n$ converges.
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0@Alex: See, for example, [here](http://www.mathscoop.com/calculus/infinite-sequences-and-series/limit-comparison-test.php). – 2012-12-10
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Hints:
1) Convergence of an infinite sum implies its terms tend to $0$.
2) If $ |a_n| \le 1$, then $ a_n^2\le |a_n|$.
3) Recall the Comparision Test for infinite sums.
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1@Neophyte Eventually, |a_n|<1. This so since the terms of a convergent series must tend to zero. – 2016-06-24