$$\sum \_{n=1}^{\infty }\frac{1}{n\left(1 +\frac{1}{2}+...+\frac{1}{n}\right)}$$

I am trying to use limit comparison test to test this series,

Let $a\_n = \frac{1}{n\left(1 +\frac{1}{2}+\ldots+\frac{1}{n}\right)}$ and $b\_n = \frac{1}{n}$

$\lim\_\limits{{n\to \infty }}\left(\frac{a\_n}{b\_n}\right)$

$=\lim \_\limits{{n\to \infty }}\left(\frac{\frac{1}{n\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)}}{\frac{1}{n}}\right)$

$=\lim \_\limits{{n\to \infty }}\left(\frac{1}{\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)}\right)$

This limit will always larger than $0$, therefore 

$\sum \_{n=1}^{\infty }\:\frac{1}{n}$ diverges, $\sum \_{n=1}^{\infty }\frac{1}{n\left(1 +\frac{1}{2}+\ldots+\frac{1}{n}\right)}$ diverges.

Is this working valid or there is a better solution for this question?