Does the series $\sum\limits_{n=1}^{\infty}\frac{\ln(\sqrt{n})}{n}$ converge or diverge?

Question

How to determine does this series converge or diverge? I have tried d'Alembert's ratio test but in the limit I get $1$. I suppose I should compare it with some other series, but I can't figure out with which one to compare to.

$$\sum\limits_{n=1}^{\infty}\dfrac{\ln(\sqrt{n})}{n}$$

Answer 1

It diverges:

$\sum\limits_{n=1}^{\infty}\dfrac{\ln(\sqrt{n})}{n} = \sum\limits_{n=1}^{\infty}\dfrac{\ln(n^{\frac{1}{2}})}{n} = \sum\limits_{n=1}^{\infty}\dfrac{\frac{1}{2}\ln(n)}{n} = \frac{1}{2}\sum\limits_{n=1}^{\infty}\dfrac{\ln(n)}{n}\geq \frac{1}{2}\sum\limits_{n=1}^{\infty}\dfrac{1}{n} = \infty$

Answer 2

Hint

$$\frac{\ln\left(\sqrt n\right)}{n}\ge \frac1n\quad\text{for n large enough}$$