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 .

Using $GM\ge HM$ $$\left( 1\cdot\frac{1}{2}\cdot \frac{1}{3}\cdots \frac{1}{n}\right)^{\frac{1}{n}}\ge \frac{n}{\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}}$$ so that $$\frac{1}{n}\cdot\frac{1}{1+\frac{1}{2}+ \frac{1}{3}+\cdots \frac{1}{n}}\ge (n!)^{\frac{1}{n}}$$ and as $(n!)^{\frac{1}{n}}>1 \forall n\ge2$ we see that $\sum (n!)^{1/n}$ cannot converge and thus $$\sum \frac{1}{n\left(1+\frac {1} {2}+\frac {1} {3}+ \cdots+\frac {1} {n}\right)}$$ cannot converge.
