If $\sum_{m=0}^{\infty} b_m$ is conditionally convergent, then is $\sum_{m=0}^{\infty} m^2b_m$ divergent? JUSTIFY

An example of conditionally convergent series is $\sum_{m=0}^{\infty} (-1)^m/\sqrt{m+1}$ and multiplying by $m^2$ it is divergent.

My conclusion is that it diverges. But what will the "justify" be?

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    $\begingroup$ if $\sum m^2 b_m$ converges, then $m^2 b_m$ is bounded. this implies $\sum |b_m|$ is bounded above by... $\endgroup$ – achille hui Dec 2 '16 at 14:10
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    $\begingroup$ You can't just show one example. You have to find either an example where the series over $m^2 a_m$ converges or a justification for why the series will always diverge. $\endgroup$ – Dominik Dec 2 '16 at 14:10

Suppose that $\sum_{m=0}^{\infty} m^2b_m$ is convergent. Hence $m^2b_m \to 0$ for $m \to \infty$. Then there is $N \in \mathbb N$ such that $m^2|b_m| \le 1$ for $m>N$. Your turn !

  • $\begingroup$ So then the series is absolutely convergent? That's kind of a strange proof. $\endgroup$ – Dubstep365 Dec 2 '16 at 17:22

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