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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?

2 Answers 2

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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.2012-12-10
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    @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.2012-12-10
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    @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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    hello can you explain how did you get $|a_n|<1$? did you use the geometric criterion? thnk you.2016-06-24
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    @Neophyte Eventually, $|a_n|<1$. This so since the terms of a convergent series must tend to zero.2016-06-24