# Convergence of $\sum\frac {(-1)^n}n + \frac 1{n\log n}$

I want to know whether $$\sum\frac {(-1)^n}n + \frac 1{n\log n}$$ converges.
My try was to use the fact that the convergence of $$\sum_{n=0}^\infty a_n$$ implies $$\lim _{n\to \infty} \frac {a_1+2a_2+...+na_n}n=0$$.
letting $$a_n=\frac {(-1)^n}n + \frac 1{n\log n}$$ ($$n=2,3,4...$$) and $$a_0=a_1=0$$, $${a_1+2a_2+...+na_{n}}=\frac{1+(-1)^{n}}2+\sum_{k=2}^n \frac 1{\log k}$$ $$\lim _{n\to \infty} \frac {a_1+2a_2+...+na_n}n=0 \iff \lim_{n\to\infty} \frac 1n\sum_{k=2}^n \frac 1{\log k}=0$$ I'm stuck right here.
How can I evaluate that limit? or is there another way to approach?

Note: the sum must start at some $$n_0>1$$ since $$\log(1)=0.$$