Converge / Diverge $\sum_{n=1}^{\infty}\big( \frac{1}{\sqrt[n]{n!}}\big)^2$ $$\sum_{n=1}^{\infty}\big( \frac{1}{\sqrt[n]{n!}}\big)^2$$
$$2^n\leq n!$$
As $2\cdot 2 \cdots2\leq 1 \cdot 2 \cdots n!$
Therefore:
$$2\leq \sqrt[n]{n!}$$
$$2^2\leq \sqrt[n]{n!}^2$$
So:
$$\frac{1}{\sqrt[n]{n!}^2}\leq \frac{1}{4}$$
$$\sum_{n=1}^{\infty}\frac{1}{\sqrt[n]{n!}^2}\leq \sum_{n=1}^{\infty}\frac{1}{4}=\frac{1}{4}\cdot \frac{1}{1-\frac{1}{4}}=\frac{1}{3}$$
Is it correct?

Since
$$
e^n  = 1 + n +  \cdots  + \frac{{n^n }}{{n!}} +  \cdots  > \frac{{n^n }}{{n!}},
$$
it holds that
$$
\sum\limits_{n = 1}^\infty  {\left( {\frac{1}{{\sqrt[n]{{n!}}}}} \right)^2 }  < \sum\limits_{n = 1}^\infty  {\frac{{e^2 }}{{n^2 }}}  = \frac{{(\pi e)^2 }}{6}.
$$
