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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?

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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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