I am given the series $\Sigma_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$, where $a_n \geq 0 \forall n$ , and $\Sigma_1^{\infty} a_n $ converges.

I was advised to expand $|\sqrt{a_n} - \frac{1}{n}|^2 $ . Doing this gives me expressions like $-\frac{2 \sqrt{a}}{n}+a+\frac{1}{n^2}$ , but I don't see how this is useful.


marked as duplicate by user940, user228113, user147263, Shailesh, Silvia Ghinassi Feb 20 '16 at 1:48

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  • $\begingroup$ Perhaps you should look into the Direct Comparison Test? $\endgroup$ – vrugtehagel Feb 19 '16 at 14:13
  • 4
    $\begingroup$ Cauchy-Schwarz: $$\left(\sum_n\frac{\sqrt{a_n}}n\right)^2\leqslant\sum_na_n\cdot\sum_n\frac1{n^2}$$ $\endgroup$ – Did Feb 19 '16 at 14:17

Writing $(x-y)^2 = x^2 - 2xy + y^2$, we see that

$$ xy \leq \frac{1}{2} \left( x^2 + y^2 \right). $$

This implies that

$$ \frac{\sqrt{a_n}}{n} \leq \frac{1}{2} \left( a_n + \frac{1}{n^2} \right). $$

Since both $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converge, the series $\sum_{n=1}^{\infty} \frac{1}{2} \left( a_n + \frac{1}{n^2} \right)$ also converges and so by the comparison test, the series $\sum_{n=1}^{\infty} \frac{\sqrt{a_n}}{n}$ also converges.

  • $\begingroup$ Very nice indeed. +1 $\endgroup$ – DonAntonio Feb 19 '16 at 14:57

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