Let be $n > 0$ an integer.
Let be $u_n = \dfrac{(-1)^{[\ln n]}}{n}$ with $[x]$ the floor part of $x$.
Does $\sum u_n$ converges?
What I have tried: we summed over the set of integers where $[\ln n]$ was constant and got an asymptotic relation of the cardinal of a such set, eventually we deduce a new expression which is divergent. I'm not exactly convinced whether I'm allowed to do this process because the series may not be absolutely convergent.
Just note that $\sum_{j=e^k}^{e^{(k+1)-1}} \frac 1 j\geq \int_{e^{k}}^{e^{(k+1)-1}} \frac 1 x dx =\ln (e^{(k+1)}-1) - \ln (e^k) =\ln \frac {e^{(k+1)}-1} {e^k} \to 1$ as $k \to \infty$. This shows that the partial sums of the original series do not form a Cauchy sequence. Hence the series is divergent.