real analysis - Does the series $sum_{nge1}frac{lnleft(frac{n+1}nright)}{sqrt n}$ converge?

Could you please give me some hint how to decide about convergence of the series
$\sum_{n\ge1}\frac{ln\left(\frac{n+1}n\right)}{\sqrt n}$ ?

I tried using comparison test:
$\frac{ln\left(\frac{n+1}n\right)}{\sqrt n}\ge \frac {ln\left(\frac1n\right)}{\sqrt n}=-\frac {ln(n)}{\sqrt n}$.
Series $\sum_{n\ge1}\frac{ln(n)}{\sqrt n}$ diverges by integral test, but for comparison test all compared sequenced must be non-negative and $ln\left(\frac1n\right)\le0$ for all n.

Thanks.

Answer

Yes it does using the asymptotic comparison:

$$\frac{\ln\left(\frac{n+1}n\right)}{\sqrt n}\sim_\infty\frac{1}{n\sqrt n}$$

Remark For the comparison test the general term of the series must have an unchanged sign i.e. not alternating series (no matter positive or negative).