convergence of $\sum_{n=1}^\infty \frac{1}{(n!)^{(2/n)}}$ Show that the following series converges or diverges.
$\sum_{n=1}^\infty \frac{1}{(n!)^{(2/n)}}$
Hint: Note that $n! \geq p^{n-p+1}$ for all $i \leq p \leq n$.
I solved this by splitting into $3$ series $A,B,C$ where $A$ will be the series for those terms for which $3$ divides $n$: i.e. $3, 6, 9...$.
$B$ will be the series for those terms for which $3$ divides $n+1$, i.e. $4, 7, 10,...$, and $C$ will be will be the series for those terms for which $3$ divides $n+2$.
Using the hint, for $A$ choose $p = (n/3 +1)$, for $B$ choose $p = (n+1)/3 +1$, for $C$ choose $p = (n+2)/3 +1$.
Each of these series $A, B, C$ converges; it should be like a $p$-series with $p = 4/3$ (not to confuse this $p$ with the previous $p$). I believe this is correct.
My question is does anyone see a cleaner/simpler way to do this?
EDIT: Thanks for the answers so far, but is there a cleaner, simpler answer based on the Above Hint?
 A: Use Stirling’s formula $n!\sim\sqrt{2\pi n}(n/e)^n$.
We get
$$(n!)^{2/n}\sim(2\pi n)^{1/n}\left(\frac ne\right)^2\sim\left(\frac ne\right)^2,\qquad n\to\infty.$$
Since $\sum\frac1{n^2}$ converges, our series converges too.
A: You can also do this via the AM-GM inequality, although it doesn't use the hint that $p^{n-p+1}\leq n!$.
AM-GM states that $$\sqrt[n]{a_1\dots a_n}\leq\frac{a_1+\dots+a_n}{n}$$  In your case, $a_j=j$; then $$\frac{1}{n!^{2/n}}=\left(\sqrt[n]{1\left(\frac{1}{2}\right)\dots\left(\frac{1}{n}\right)}\right)^2\leq\left(\frac{1+\dots+\frac{1}{n}}{n}\right)^2\leq\frac{(\ln(n)+1)^2}{n^2}$$  Thus $$\sum_n{\frac{1}{n!^{2/n}}}\leq\sum_n{\frac{(\ln(n)+1)^2}{n^2}}$$ and the latter is convergent by noting, e.g., that each summand is $o(n^{-2+\epsilon})$.
A: For every positive integer $n$ we have
$$e^n=\sum_{p=0}^{+\infty}\frac{n^p}{p!}>\frac{n^n}{n!}$$
i.e.
$$\frac{1}{n!}<\left(\frac{e}{n}\right)^n$$
and thus
$$\frac{1}{(n!)^{2/n}}<\frac{e^2}{n^2}$$
defines a convergent series.
A: $$a_n=\frac{1}{(n!)^{(2/n)}}\implies \log(a_n)=-\frac 2 n \log(n!)$$ Using Stirling approximation
$$\log(a_n)=2(1- \log (n))-\frac{\log (2 \pi  n)}{n}-\frac{1}{6
   n^2}+O\left(\frac{1}{n^4}\right)$$ Apply it for $\log(a_{n+1})$ and continue with Taylor expansions to get
$$\log(a_{n+1})-\log(a_n)=-\frac{2}{n}+\frac{\log (2 \pi  n)}{n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} } { a_n}=e^{\log(a_{n+1})-\log(a_n)}\sim e^{-\frac{2}{n}}\quad \to ~~~ 0$$ So, convergence
