$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$ I have to study the character of this series 
$$\sum_{n=1}^\infty (n!)^{-\frac{1}{n}}$$
and I tried with the ratio test:
$ \frac{a_{n+1}}{a_n}= \frac{((n+1)!)^{-\frac{1}{n+1}}}{(n!)^{-\frac{1}{n}}}=
\frac{(n!)^{\frac{1}{n}}}{((n+1)!)^{\frac{1}{n+1}}}=
\frac{(n!)^{\frac{1}{n}}}{((n+1) \star n!)^{\frac{1}{n+1}}}=
\frac{(n!)^{\frac{1}{n}-\frac{1}{n+1}}}{((n+1) )^{\frac{1}{n+1}}} \sim (n!)^{\frac{1}{n}-\frac{1}{n+1}}=
(n!)^{\frac{1}{n(n+1)}}>(n(n+1))^{\frac{1}{n(n+1)}} \sim 1$
So the limit is larger than 1 and the given series diverges.
Is it right?

$\frac{1}{n}=\frac{1}{n^{n*\frac{1}{n}}}<\frac{1}{n!^{\frac{1}{n}}}$
Applying the comparison test and knowing that the series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, also our series diverges
