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! 

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.
