Convergence/absolute convergence of the series $\sum_{n=2}^{\infty} (-1)^{n+1}\frac{\ln (n)}{n}$

Question

Does the series $\sum_{n=2}^{\infty} (-1)^{n+1}\frac{\ln (n)}{n}$ converge? If so, does it converge absolutely?

My attempt:

Because $$\left(\frac{\ln x}{x}\right)' = \frac{1-\ln(x)}{x²} > 0$$ whenever $x > e$, and because $\ln(n)/n \to 0$, the Leibniz criterion tells us that the series converges.

Because $$\int_{1}^{+ \infty} \frac{\ln(x)}{x}dx = \lim_{k \to \infty} 1/2\ln^2(k) = + \infty$$

the series does not converge absolutely. Is this correct?

Answer 1

More simply, without calculating the integral, it is because $\dfrac{\ln n}n>\dfrac 1n$ if $n>2$, and the harmonic series diverges.

Answer 2

You can show with $\displaystyle S_{2n}=\sum_{k=1}^n\bigg(\dfrac{\ln(2k+1)}{2k+1}-\dfrac{\ln2k}{2k}\bigg)$ that $\bigg(\dfrac{\ln(2n+1)}{2n+1}-\dfrac{\ln2n}{2n}\bigg)=o(n^{-3/2})$ so $S_{2n}$ converges.