# If $\lim\limits_{x\to\infty}\frac{n}{u_n}=1$, what can we say about $\sum\limits_{n=1}^{\infty}(-1)^n \frac{1}{u_n}$?

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!

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

If $\dfrac1{u_n} = \dfrac1{n}+(-1)^n\dfrac1{n\ln(n)}$, then
$\begin{array}\\ \sum_{n=1}^M (-1)^n\dfrac1{u_n} &=\sum_{n=1}^M (-1)^n(\dfrac1{n}+(-1)^n\dfrac1{n\ln(n)})\\ &=\sum_{n=1}^M (-1)^n\dfrac1{n}+\sum_{n=1}^M\dfrac1{n\ln(n)}\\ \end{array}$
• Thank you. I think your method is a specific example of the way Daniel Fisher proposed in the comment $\frac{n}{u_n} = 1 + (-1)^n\varepsilon_n$. – M. Chen Sep 29 '17 at 14:16