# Evaluate in closed form: $\sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$

Evaluate in closed form: $$\sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$$

I tried to use same method for similar question and two variables.Was not able to get the final answer. Question proposed by Jalil Hajimir -- Infinite Series $\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{m!\:n!}{(m+n+2)!}$

In the book (Almost) Impossible Integrals, Sums, and Series, page $$533$$, the result in $$(6.291)$$, it is easily shown by transforming the Gamma summand into a summand involving a product of Beta function (which we can further deal with using the integral representation of Beta function) that $$\sum_{i=1}^{\infty}\left( \sum_{j=1}^{ \infty} \frac{\Gamma(i)\Gamma(j)\Gamma(x)}{\Gamma(i+j+x)}\right)=\frac{1}{2}\left(\psi^{(1)}\left(\frac{x}{2}\right)-\psi^{(1)}\left(\frac{1+x}{2}\right)\right).$$

Now, replacing $$x$$ by $$k$$, considering the summation from $$k=1$$ to $$\infty$$, passing from Gamma function to factorials and reindexing the series, we get in the left-hand side the series the Op is interested in, and in the right-hand side a simple telescoping sum involving Trigamma function. The answer is indeed $$\pi^2/4$$.

Playing with Mathematica a bit and you will see that the sum is $$S(3)=\sum_{p=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+3)!} = \frac{1}{12} \left(3 \zeta (3)+2 \pi ^2 \log (2)\right)+ \sum _{n=1}^{\infty } \left(\frac{\psi ^{(1)}(n)}{4 n (2 n-1)}-\frac{\psi ^{(1)}\left(n+\frac{1}{2}\right)}{4 n (2 n-1)}\right)$$ where $$\psi$$ is the digamma function. It does not look like it can be further simplified.

Curiously though, a slightly different sum do has a closed form $$S(2)=\sum_{p=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+2)!} =\pi^2/4$$ To see this, one can check $$S(2) =\sum_{p=0}^\infty \sum_{m=0}^\infty\frac{m! p!}{(m+p+1)^2 (m+p)!} =\sum_{p=0}^\infty\frac{\, _3F_2(1,1,p+1;p+2,p+2;1)}{(p+1)^2} =:\sum_{p=0}^\infty a_p$$

Mathematica cannot simplify this anymore, but look at the first few terms $$\frac{\pi ^2}{6},\frac{1}{6} \left(12-\pi ^2\right),\frac{1}{6} \left(\pi ^2-9\right),\frac{1}{18} \left(31-3 \pi ^2\right),\frac{1}{72} \left(12 \pi ^2-115\right),\frac{3019-300 \pi ^2}{1800},\frac{1}{600} \left(100 \pi ^2-973\right),\frac{48877-4900 \pi ^2}{29400}$$ The pattern is quite obvious, we should have $$a_{2p}+a_{2p+1}= \frac{2}{(2 p+1)^2}, \quad p \ge 0.$$ So $$S(2)=\sum_{p \ge 0} \frac{2}{(2 p+1)^2} = \pi^2/4.$$

• Original question is not 3 it is 2 – Keon Dec 10 '19 at 23:03
• fWhen it is 2 it has a closed form. – Keon Dec 10 '19 at 23:07
• @Keon The question and the title does not match. – ablmf Dec 10 '19 at 23:10
• It is fixed. Now both of them are 2 – Keon Dec 10 '19 at 23:13