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This question already has an answer here:

Is possible to find series that: $\sum_{n=1}^\infty a_n$ is converges but $\sum_{n=1}^\infty a_n^3$ is diverges

I thought about something with $(-1)^n$ and Leibniz criterion but don't have idea.

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marked as duplicate by Andrés E. Caicedo, user61527, TZakrevskiy, Dietrich Burde, Hanul Jeon Nov 30 '13 at 0:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Yes, this is possible. $\endgroup$ – Andrés E. Caicedo Nov 29 '13 at 17:43
  • $\begingroup$ While this is a not very useful comment (in how it doesn't give a concrete answer), the answer is that yes, it is possible. If this question is from the same place I encountered it, then the lecturer's comment on this question may be entertaining: "Going off the answers to the prior questions on the [homework] sheet, the answer is probably yes." $\endgroup$ – Andrew D Nov 29 '13 at 17:46
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Let $j=e^{2\pi i/3}$.

$\sum \cfrac{j^k}{k}$ converges but $\sum \cfrac{1}{k}$ diverges.

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  • $\begingroup$ The question is tagged real-analysis... $\endgroup$ – TonyK Nov 29 '13 at 19:59
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Maybe: $$\left(1,\frac{-1}{2} ,\frac{-1}{2} ,\frac{1}{\sqrt[3]{2}} ,\frac{1}{2}\cdot \frac{-1}{\sqrt[3]{2}} ,\frac{1}{2}\cdot \frac{-1}{\sqrt[3]{2}} ,..., \frac{1}{\sqrt[3]{n}} ,\frac{1}{2}\cdot \frac{-1}{\sqrt[3]{n}}, \frac{1}{2}\cdot \frac{-1}{\sqrt[3]{n}} ,\frac{1}{\sqrt[3]{n+1}} ,\frac{1}{2}\cdot \frac{-1}{\sqrt[3]{n+1}}, \frac{1}{2}\cdot \frac{-1}{\sqrt[3]{n+1}} ,...\right)$$

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