Convergence of $\sum \frac{{a}_{n}}{\sqrt{n}}$ given that $\sum {a}_{n}^{2}$ converges

Joel French
2022-07-13
Answered

Convergence of $\sum \frac{{a}_{n}}{\sqrt{n}}$ given that $\sum {a}_{n}^{2}$ converges

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Dayana Zuniga

Answered 2022-07-14
Author has **16** answers

Let ${a}_{n}=\frac{1}{\sqrt{n}\mathrm{log}n}$. Then $\sum _{n=2}^{\mathrm{\infty}}{a}_{n}^{2}=\sum _{n=2}^{\mathrm{\infty}}\frac{1}{n{\mathrm{log}}^{2}n}$ which converges by Cauchy's condensation test, but

$\sum _{n=2}^{\mathrm{\infty}}\frac{{a}_{n}}{\sqrt{n}}=\sum _{n=2}^{\mathrm{\infty}}\frac{1}{n\mathrm{log}n}$

and this diverges, again by Cauchy's condensation test.

$\sum _{n=2}^{\mathrm{\infty}}\frac{{a}_{n}}{\sqrt{n}}=\sum _{n=2}^{\mathrm{\infty}}\frac{1}{n\mathrm{log}n}$

and this diverges, again by Cauchy's condensation test.

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