Check the convergence of $\sum_{n=2}^{\infty} \frac{(-1)^n}{n\cdot \ln^{7/5}(n)}$ 
I want to check the convergence of this series
  $$\sum_{n=2}^{\infty} \frac{(-1)^n}{n\cdot \ln^{7/5}(n)}.$$

When I have this kind of series I pick the positive one and check the tests on it.
$$\sum_{n=2}^{\infty} \frac{1}{n\cdot \ln^{7/5}(n)}$$
I can say that 
$$n>\ln(n) \Rightarrow \frac{1}{n} < \frac{1}{\ln(n)}$$

My question is if I have chosen the right series for this test.

 A: Hint: Use alternating series test.
A: In fact, here your series converges absolutely since we know that
$$ \sum_{n=2}^{+\infty} \frac{1}{n^\alpha (\ln n)^{\beta}}<+\infty \Leftrightarrow \left[(\alpha>1)\text{ or }(\alpha=1\text{ and }\beta >1)\right]$$
You can prove this with integral test. And conclude since the absolute convergence implies the convergence.
A: Edited : thanks to Did
There is no need for any test, the terms of series tend to $0$ monotonically alternating sign, that is sufficient that series converges. (There was nothing regrading the absolute convergence in the question, so why bother with it?)
More over the limit of the series L is :
$$\frac{(-1)^2}{2\cdot \ln^{7/5}(2)}+\frac{(-1)^3}{3\cdot \ln^{7/5}(3)}<L<\frac{(-1)^2}{2\cdot \ln^{7/5}(2)} $$
A: Hint:
For $a_n=\frac{1}{n^\alpha \ln^\beta n}$ ($n\geq 2$), the positive series $\sum a_n$ converges if


*

*$\alpha > 1$; or

*$\alpha = 1$ and $\beta > 1$


(and diverges otherwise.)
This'll allow you to see if your series converges absolutely.
