Convergence/Divergence of some series If I have $\sum_{n=2}^{\infty} \frac {1}{n\log n}$ and want to prove that it diverges, can I use following?
$$\frac {1}{n\log n} \lt \frac {1}{n}$$


*

*$\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of $\frac 1 n$ equals to zero so the comparasion I think isn't useful.

*Or can I say it diverges because $\sum_{n=1}^\infty \frac 1{\log n}$ also diverges?
And If I have $\sum_{n=2}^\infty \frac {(-1)^n}{(n)}$ I can use the Leibniz rule. $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that $\sum_{n=1}^\infty \frac 1 n$ has an infinite sum, but what to do in this case?

Since
$$
\sum_{n=2}^\infty\frac{2^n}{2^n n\log(2)}
$$
diverges by comparison to the Harmonic Series, the Cauchy-Condensation Test says that
$$
\sum_{n=2}^\infty\frac1{n\log(n)}
$$
also diverges.
