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December 17th, 2018, 04:51 PM   #1
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Convergence proof

How to show that if $\displaystyle \sum_{n=1}^\infty x^2_n $ converges, then $\displaystyle \sum_{n=1}^\infty \frac{x_n}{n} $ also converges?

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December 17th, 2018, 08:26 PM   #2
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For any $n$, you have $(x_n - \frac{1}{n})^2 > 0$ which implies the bound $\frac{2x_n}{n} \leq x_n^2 + \frac{1}{n^2}$.
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December 18th, 2018, 04:33 AM   #3
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For positive ${x_n}$.
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December 18th, 2018, 05:48 AM   #4
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$\displaystyle \left(\left|x_n\right| - \frac1n\right)^2 \geqslant 0 \implies \left|\frac{2x_n}{n}\right| \leqslant x_n^2 + \frac{1}{n^2} \implies \sum_{n=1}^\infty \frac{x_n}{n}$ is absol.utely convergent.
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