calculus - Is $sumlimits_{n=2}^{infty} frac{(-1)^n n^2}{( ln(n))^n}$ divergent?

The series in question is $\sum\limits_{n=2}^{\infty}\frac{(-1)^n n^2}{( \ln(n))^n}$
I think this diverges. So using the divergence test, I am trying to show that limit of the general term of the series is not 0.
By a few algebraic manipulations we get :

• $\lim (-1)^n(n^2)( \ln(n))^{-n}$


• $\lim (-1)^n(n)/ \ln(n)$

I am not sure how to proceed here. If you get rid of the alternating, the limit is $\infty$.
However, I am not sure how to manipulate this to show that (I suspect) the limit does not exist.

Answer

Applying the root test you can see that the series converges.

$$L = \lim_{n\to\infty} |a_{n}|^{1/n} .$$

If $L<1$, then the series is absolutely convergent.

If $L>1$ the series is divergent.

If $L=1$ then the test is inconclusive. However, if $|a_{n}|^{1/n}\geq 1$ for infinitely many distinct values of n, then the series $\sum_{n=1}^{\infty} a_{n}$ diverges.

In your series, you have

$$L = \lim_{n\to\infty}\bigg| \frac{(-1)^{n}n^{2}}{\text{ln}^{n}(n)}\bigg|^{\frac{1}{n}}$$

Then $L=\lim \limits_{n\to\infty}\bigg|\displaystyle\frac{(-1)n^\frac{2}{n}}{\text{ln}(n)}\bigg|=0$ (because $\lim \limits_{n\to\infty}\ln(n)=\infty$).

Therefore, $L<1$ and the series convergences.