# If $\sum a_n^2 n^2$ converges then $\sum |a_n|$ converges

Let's suppose that $(a_n)$ is a sequence so that $\sum a_n^2 n^2$ converges, so I have to prove that $\sum |a_n|$ converges.

By the Cauchy criterion we have that, since $\sum a_n^2 n^2$ converges if $\varepsilon > 0$ then there is a $k\in\mathbb N$ so for all $m,n$ naturals that satisfy $m > n \geq k$: $$|n^2 a_n^2 + ... + m^2 a_m^2| < \varepsilon$$ but I can't get $| |a_n| + ... + |a_m| | < \varepsilon$ to show that $\sum |a_n|$ converges. Could you give me a hint? Is there any way to prove this without resorting Cauchy criterion. Thanks!

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Try using this inequality: $$\lvert a_n\rvert + \lvert a_{n+1}\rvert +\ldots+\lvert a_m\rvert \le \sqrt{m-n}\sqrt{a_n^2+\ldots+a_m^2}.$$ –  Giuseppe Negro Feb 7 '13 at 1:52
Use the inequality $pq\le{1\over 2}(p^2+q^2)$ with $p=1/n$ and $q=n|a_n|$ (or better, argue as in André's answer). –  David Mitra Feb 7 '13 at 2:32

One can do it without using the Cauchy Criterion. The Comparison Test is enough.

Suppose that $a_n \ne 0$ and $|a_n| \gt n^2a_n^2$. Then $|a_n|\lt \frac{1}{n^2}$.

For all other $n$, we have $|a_n|\le n^2a_n^2$.

Thus for all $n$ we have $|a_n|\le \frac{1}{n^2}+n^2a_n^2$.

The series $\sum \frac{1}{n^2}$ converges, as does, by assumption, $\sum n^2a_n^2$. Thus the series $\sum\left(\frac{1}{n^2}+n^2a_n^2 \right)$ converges.

By Comparison, $\sum|a_n|$ converges.

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Hint: $\sum_{n=1}^\infty 1/n^2$ converges. Use the Cauchy–Schwarz inequality.

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Yes, way to go. +1 –  1015 Feb 7 '13 at 2:15