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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.

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    What have you attempted so far to solve this problem? – 2017-01-03
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    The question is not very clear: are you just wondering about convergence or do you need to find a closed formula for such series? – 2017-01-03
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    @wythagoras I tried the d'Alembert Criteria but it doesn't simplify itself enough to prove the convergence.. – 2017-01-03
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    @JackD'Aurizio yea, I'm looking for the general expression that proves that this serie converge, cause the only way I've founded is to try with numbers bigger than 2.. so not really legit – 2017-01-03

3 Answers 3

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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.

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    Ok, thank you for your answer! :) could you please still explain me why it's only if d is equal or bigger than 2? Cause if d=1, every others terms are smaller than one and 1/d is equal to one, so by the d'Alembert ratio, should be convergent too?! – 2017-01-03
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    @JohnAdams In the case $d=1$, there is only the term that is equal to $\frac{1}{d}=1$, and d'Alembert ratio test requires the ratio to be strictly smaller than 1. If $d=1$, each term in the sum is in fact equal to 1. – 2017-01-03
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    Oh yea...obvious, thank you very much! – 2017-01-03
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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:

$$\begin{eqnarray*} S_d = \int_{(0,+\infty)^d}e^{-\sum x_k}\sum_{n\geq 0}\frac{(x_1\cdot\ldots\cdot x_d)^n}{(nd)!}\,d\mu &\leq& \int_{(0,+\infty)^d}e^{-\sum x_k}\sum_{n\geq 0}\frac{(\sum x_k)^{dn}}{d^{nd}(nd)!}\,d\mu\\&\leq&\int_{(0,+\infty)^d}e^{-\sum x_k}\sum_{m\geq 0}\frac{(\sum x_k)^{m}}{d^{m}m!}\,d\mu\\&=&\int_{(0,+\infty)^d}\exp\left[-\left(1-\frac{1}{d}\right)\sum x_k\right]\,d\mu\\&=&\frac{1}{\left(1-\frac{1}{d}\right)^d}. \end{eqnarray*}$$

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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.