Convergence of $ \sum_{n=1} ^\infty \frac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$ Convergence of $$ \sum_{n=1} ^\infty \dfrac {1}{n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} )}$$
Attempt: I believe not a nice attempt: $ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n( 1+1+\cdots+1 )$ 
$\implies n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) \leq n^2$
$\implies \dfrac {1}{ n(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n} ) } \geq \dfrac {1}{n^2}$
I guess this result is of not much use. Can somebody please tell me a direction to move ahead in this problem?
Thank you very much for your help .

Hint. Recall that, as $n \to +\infty$, we have
$$1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\sim \ln n$$
then
$$n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)\sim n\ln n$$
and
$$
\sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)} \sim \sum \frac{1}{n \ln n},
$$ your initial series is then divergent as is the latter series.
