Find a positive sequence $\{a_n\}$ such that $\sum_{n}\frac{1}{a_n}$ and $\sum _{n}\frac{a_n}{n^2}$ converge My Try :


So we have :

$$\sum _{n=1}^{\infty}\dfrac{1}{n^{e-1}} \ \ \ \text{ is converges}$$

But :

$$\sum _{n=1}^{\infty}\dfrac{n^{e-1}}{n^2} \ \ \ \text{ is Divergent :(}$$

Please help me !

  • 2
    $\begingroup$ You can’t do it with $n^k$, because either $k\le 1$, and the first series diverges, or $2-k\le 1$, and the second series diverges. $\endgroup$ – G Tony Jacobs Nov 26 '17 at 14:19
  • 2
    $\begingroup$ Hint: Use the inequality $2\sqrt{uv}\leq u+v$ ($u,v\geq 0$) for $u=1/a_n$ and $v=a_n/n^2$. $\endgroup$ – Kelenner Nov 26 '17 at 14:23
  • $\begingroup$ @Kelenner: Thank you for spotting the error. This necessary condition is only true for a positive and monotonic decreasing series. $\endgroup$ – MrYouMath Nov 26 '17 at 14:45

There is no such sequence. If there were, by Cauchy-Schwarz $$\sum_{k=1}^N \frac 1k=\sum_{k=1}^N \frac 1{\sqrt{a_k}} \frac{\sqrt{a_k}}{k}\leq \sqrt{\sum_{k=1}^N \frac 1{a_k}} \sqrt{\sum_{k=1}^N \frac{a_k}{k^2}}\leq \sqrt{\sum_{k=1}^\infty \frac 1{a_k}} \sqrt{\sum_{k=1}^\infty \frac{a_k}{k^2}} $$

This implies convergence of the harmonic series, a contradiction.

  • $\begingroup$ Nice solution. Thank you for sharing. $\endgroup$ – MrYouMath Nov 26 '17 at 14:46

Assume that both series converges. Since $a_{n}$ is a sequence of positive real numbers then the convergence is also absolute, so we can rearrange the series. Then we can observe that $$\sum_{n\leq N}\left(\frac{a_{n}}{n^{2}}+\frac{1}{a_{n}}\right)=\sum_{n\leq N}\left(\frac{a_{n}^{2}+n^{2}}{n^{2}a_{n}}\right)\geq\sum_{n\leq N}\left(\frac{2a_{n}n}{n^{2}a_{n}}\right)=\sum_{n\leq N}\frac{2}{n}\tag{1}$$ where the inequality follows from the AM-GM inequality. Hence taking the limit $N\rightarrow \infty$ we have a contradiction.

  • $\begingroup$ This answer would be improved by adding some words explaining the significance of this calculation. :) $\endgroup$ – G Tony Jacobs Nov 26 '17 at 16:02
  • $\begingroup$ @GTonyJacobs You are absolutely right but yesterday I was in hurry :D $\endgroup$ – Marco Cantarini Nov 27 '17 at 8:04
  • $\begingroup$ @MarcoCantarini . why $\sum_{n\leq N}\left(\frac{a_{n}}{n^{2}}+\frac{1}{a_{n}}\right)$ is converges ? $\endgroup$ – Almot1960 Nov 27 '17 at 8:59
  • $\begingroup$ @Almot1960 I assumed that both $\sum_{n\geq1}\frac{1}{a_{n}}$ and $\sum_{n\geq1}\frac{a_{n}}{n^{2}}$ converges. Then, due to the fact that $a_{n}>0$ and so the absolute convergence, $\sum_{n\geq1}(\frac{a_{n}}{n^{2}}+\frac{1}{a_{n}})$ converges. But if we take the partial sum we have an inequality that contadicts this convergence. $\endgroup$ – Marco Cantarini Nov 27 '17 at 9:15

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