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?
$$\frac {1}{n\log n} \lt \frac {1}{n}$$


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*$\sum_{n=1}^\infty \frac 1 n$ 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?

Bertrand's series $\displaystyle \sum\limits_{n\ge 2}\frac 1{n^\alpha\ln(n)^\beta}$ converges only for $(\alpha>1)$ or $(\alpha=1,\beta>1)$.
This is generally proved using the comparison to an integral as other answers have shown.
But in general for series of the kind $\displaystyle \sum\limits_{n\ge 2}\frac 1{n^{\alpha_0}\ln(n)^{\alpha_1}\ln(\ln(n))^{\alpha_2}\ln(\ln(\ln(n)))^{\alpha_3}...}$ 
Which all share the same property of converging only if $\alpha_i=1$ for $i<k$ and $\alpha_k>1$ (with $k$ being the last one), you can use the condensation Cauchy test.
The criterion is : For $(a_n)_n \searrow\ \ge 0$ then $\displaystyle S=\sum\limits_{n\ge 1} a_n<+\infty\iff T=\sum\limits_{n\ge 0} 2^na_{2^n}<+\infty$ 
with inequality $S\le T\le 2S$.

Applied to $\displaystyle S=\sum\limits_{n\ge 2}\frac 1{n\ln(n)}$, it gives $\displaystyle T=\sum\limits_{n\ge 1}\frac{2^n}{2^n\ln(2^n)}=\frac 1{\ln(2)}\sum\limits_{n\ge 1} \frac 1n$ which is divergent.
