How to prove the convergence of $\sum_{n\geq 0}\frac{n!^d}{(dn)!}$ for $d\geq 2$? How can we find the general formula about the convergence of the serie $\sum_{n=0}^\infty \frac{(n!)^d}{(dn)!}$ for all $d\geq2$ ?
I tried using the d'Alembert Criteria but it doesn't simplify itself enough to prove the convergence.
 A: Using d'Alembert's ratio test actually works. $$\frac{\left(\frac{((n+1)!)^d}{(d(n+1))!}\right)}{\left(\frac{(n!)^d}{(dn)!}\right)} = \frac{((n+1)!)^d}{(n!)^d} \cdot \frac{(dn)!}{(d(n+1))!} = (n+1)^d \cdot \frac{1}{(dn+1)(dn+2)\cdots(dn+d)} = \frac{(n+1)}{(dn+1)} \cdot \frac{(n+1)}{(dn+2)} \cdots \frac{(n+1)}{(dn+d)}  $$
Now the last term is $\frac1d$, and the other terms are smaller than one.
A: If $d=2$, using Euler's Beta function
$$\begin{eqnarray*} S_2 = \sum_{n\geq 0}\frac{n!^2}{(2n)!}=\sum_{n\geq 0}\frac{\Gamma(n+1)^2}{\Gamma(2n+1)}&=&\sum_{n\geq 0}(2n+1)\,B(n+1,n+1) \\&=&\int_{0}^{1}\sum_{n\geq 0}(2n+1)x^n(1-x)^n\,dx\\&=&\int_{0}^{1}\frac{1+x(1-x)}{(1-x(1-x))^2}\,dx\\&=&\int_{-1/2}^{1/2}\frac{\tfrac{5}{4}-x^2}{\left(\tfrac{3}{4}+x^2\right)^2}\,dx\\&=&4\int_{0}^{1}\frac{(5-x^2)}{(3+x^2)^2}\,dx\\&=&\color{red}{\frac{2}{27}\left(18+\pi\sqrt{3}\right)}\end{eqnarray*} $$
and in general, by setting $S_d=\sum_{n\geq 0}\frac{n!^d}{(dn)!}$ we have that the ratio between the term associated with $n=N+1$ and the term associated with $n=N$ is given by:
$$ \frac{(N+1)^d}{(dN+d)(dN+d-1)\cdot\ldots\cdot(dN+1)}\leq\frac{1}{d} $$
so the series is convergent by the ratio test/a comparison with a geometric series. An alternative is given by the AM-GM inequality:
$$\begin{eqnarray*} S_d = \int_{(0,+\infty)^d}e^{-\sum x_k}\sum_{n\geq 0}\frac{(x_1\cdot\ldots\cdot x_d)^n}{(nd)!}\,d\mu &\leq& \int_{(0,+\infty)^d}e^{-\sum x_k}\sum_{n\geq 0}\frac{(\sum x_k)^{dn}}{d^{nd}(nd)!}\,d\mu\\&\leq&\int_{(0,+\infty)^d}e^{-\sum x_k}\sum_{m\geq 0}\frac{(\sum x_k)^{m}}{d^{m}m!}\,d\mu\\&=&\int_{(0,+\infty)^d}\exp\left[-\left(1-\frac{1}{d}\right)\sum x_k\right]\,d\mu\\&=&\frac{1}{\left(1-\frac{1}{d}\right)^d}. \end{eqnarray*}$$
A: Another way for proving the convergence is using the Stirling's approximation $$\frac{n!^{d}}{\left(nd\right)!}\sim\frac{\left(2\pi\right)^{d/2}n^{d/2}n^{nd}}{e^{nd}}\frac{e^{nd}}{\left(2\pi\right)^{1/2}\left(nd\right)^{1/2}\left(nd\right)^{nd}}=\frac{\left(2\pi\right)^{d/2-1/2}n^{d/2-1/2}}{d^{nd+1/2}}$$ as $n\rightarrow \infty$ and so the convergence.
