2
$\begingroup$

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!

$\endgroup$
  • 4
    $\begingroup$ Play with $$\frac{n}{u_n} = 1 + (-1)^n\varepsilon_n. $$ $\endgroup$ – Daniel Fischer Sep 28 '17 at 13:47
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.