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?


The first converges by the alternating series test, but the second diverges using the integral test, so the sum (your series) diverges.

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

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