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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?

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marked as duplicate by Git Gud, mau, Davide Giraudo, Etienne, TooTone Mar 14 '14 at 15:17

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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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This doesn't work. For example, it's trivially true that if $\displaystyle \frac{\displaystyle\frac{1}{n!}}{\displaystyle \frac{1}{n}}\to 0$ but the Harmonic Series doesn't converge. – Alex Youcis Dec 10 '12 at 17:32
@Alex: That doesn't work as a counterexample. I'm not concluding that the series consisting of terms from the denominator converges; I'm concluding that the series consisting of terms from the numerator converges. That makes all the difference. The fact that the limit of the ratios is $0$ means that the terms in the numerator are essentially smaller than the terms in the denominator. Since the larger (denominator terms) series converges, so does the smaller (numerator terms) series. – Mike Spivey Dec 10 '12 at 17:42
@Alex: See, for example, here. – Mike Spivey Dec 10 '12 at 17:49


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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hello can you explain how did you get $|a_n|<1$? did you use the geometric criterion? thnk you. – Neophyte Jun 24 at 21:21
@Neophyte Eventually, $|a_n|<1$. This so since the terms of a convergent series must tend to zero. – David Mitra Jun 24 at 21:28

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