If $\lim\limits_{x\to \infty}\frac{n}{u_n}=1$, can we prove that series $\sum\limits_{n=1}^{\infty}(-1)^n \frac{1}{u_n}$ converges, doesn't converge or it's indefinite?

I know that $\sum\limits_{n=1}^{\infty}\frac{1}{u_n}$ diverges by the comparing theorem. But I didn't find an explicit theorem about the alternate series.

Any help will be appreciated, thank you!

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    $\begingroup$ Play with $$\frac{n}{u_n} = 1 + (-1)^n\varepsilon_n. $$ $\endgroup$ – Daniel Fischer Sep 28 '17 at 13:47
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    $\begingroup$ A series may only be convergent or non-convergent, what does it mean for a series to be indefinite? $\endgroup$ – Jack D'Aurizio Sep 28 '17 at 14:20
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    $\begingroup$ @JackD'Aurizio Given a problem of this sort, it is possible that it could converge or not, given certain sequences. $\endgroup$ – Isaac Browne Sep 28 '17 at 22:01
  • $\begingroup$ @IsaacBrowne Not = divergence. there is no third category $\endgroup$ – zhw. Sep 28 '17 at 22:10
  • $\begingroup$ @zhw Could I not say that it depends on the sequence. Here is a scenario. A random series $u_n$ may diverge or it may converge. It depends on what the sequence is. $\endgroup$ – Isaac Browne Sep 28 '17 at 22:12

If $\dfrac1{u_n} = \dfrac1{n}+(-1)^n\dfrac1{n\ln(n)} $, then

$\begin{array}\\ \sum_{n=1}^M (-1)^n\dfrac1{u_n} &=\sum_{n=1}^M (-1)^n(\dfrac1{n}+(-1)^n\dfrac1{n\ln(n)})\\ &=\sum_{n=1}^M (-1)^n\dfrac1{n}+\sum_{n=1}^M\dfrac1{n\ln(n)}\\ \end{array} $

and the first sum converges while the second sum diverges, so the sum itself diverges.

Therefore you can not say.

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  • $\begingroup$ Thank you. I think your method is a specific example of the way Daniel Fisher proposed in the comment $\frac{n}{u_n} = 1 + (-1)^n\varepsilon_n$. $\endgroup$ – M. Chen Sep 29 '17 at 14:16

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