# Triple series - evaluation delayed

Trying to figure out if the infinite triple series has a nice closed form. It seems Mathematica
is unable to help us here. Numerically, things remain the same, no response. Could you help?

N[ Sum[ n! k! m!/(n + k + m + 2)!, {n, 0, ∞}, {k, 0, ∞}, {m, 0, ∞}]]


EDIT: in 2 variables, the double series evaluates to $\pi^2/6$.

• You can do two of the sums, one at a time. Commented Dec 23, 2013 at 9:10
• The answer is $\pi^2/4$ and it can be guessed via Mathematica. Shortly, evaluating the sums one by one. Now I'm going to sleep, will write tomorrow if somebody wouldn't do it earlier. Commented Dec 23, 2013 at 20:49
• @Chris'ssis please see the answer below. Numerical checking give near enough values. Commented Dec 24, 2013 at 8:27

Doing sums one by one explicitly specifying parameters helps:

f1 = Sum[ n! k! m!/(n + k + m + 2)!, {k, 0, ∞},
Assumptions -> k ∈ Integers && m ∈ Integers && k > 0 && m > 0]


gives $$\frac{\Gamma (m+1) \Gamma (n+1)}{(m+n+1)^2 \Gamma (m+n+1)}$$

f2 = Sum[ f1, {m, 0, ∞}, Assumptions -> n ∈ Integers && n > 0]


$$\frac{\, _3F_2(1,1,n+1;n+2,n+2;1)}{(n+1)^2}$$ Mathematica doesn't evaluate the last Sum[ f2, {n, 0, ∞}], but here a very handy FindSequenceFunction can be used:

tt = Table[f2, {n, 1, 10}] // Expand


$$\left\{2-\frac{\pi ^2}{6},\frac{\pi ^2}{6}-\frac{3}{2},\frac{31}{18}-\frac{\pi ^2}{6},\frac{\pi ^2}{6}-\frac{115}{72},\frac{3019}{1800}-\frac{\pi ^2}{6},\frac{\pi ^2}{6}-\frac{973}{600},\frac{48877}{29400}-\frac{\pi ^2}{6},\frac{\pi ^2}{6}-\frac{191833}{117600},\frac{5257891}{3175200}-\frac{\pi ^2}{6},\frac{\pi ^2}{6}-\frac{5194387}{3175200}\right\}$$

f3 = FullSimplify[ FindSequenceFunction[tt, n], n ∈ Integers && n > 0]


$$\frac{1}{2} \left(\psi ^{(1)}\left(\frac{n+1}{2}\right)-\psi ^{(1)}\left(\frac{n+2}{2}\right)\right)$$ Somehow on less than ten terms in $tt$ FindSequenceFunction doesn't work here.

So

Sum[ f3, {n, 0, ∞}]


is a telescopic sum and is equal to the first term $\frac12\psi ^{(1)}\left(\frac{1}{2}\right)=\frac{\pi ^2}{4}$.

• This seems too nice to be true ... Commented Dec 24, 2013 at 12:54
• FindSequenceFunction is just a guess, since it operates on a finite portion of the sequence; is that not correct? Commented Dec 27, 2013 at 8:50
• @robjohn yes. $~$ Commented Dec 27, 2013 at 9:01
• Nonetheless, nice answer! (+1) Commented Dec 27, 2013 at 9:04

Note that \begin{align} &\frac{k!}{(n+k+1)!}-\frac{(k+1)!}{(n+k+2)!}\\ &=(n+k+2)\frac{k!}{(n+k+2)!}-(k+1)\frac{k!}{(n+k+2)!}\\ &=(n+1)\frac{k!}{(n+k+2)!}\tag{1} \end{align} Sum $(1)$ for $k\ge0$ and divide by $n+1$ to get $$\sum_{k=0}^\infty\frac{k!}{(n+k+2)!}=\frac1{(n+1)(n+1)!}\tag{2}$$ Applying $(2)$ to the triple sum yields \begin{align} \sum_{n=0}^\infty\sum_{m=0}^\infty\sum_{k=0}^\infty\frac{n!m!k!}{(n+m+k+2)!} &=\sum_{n=0}^\infty\sum_{m=0}^\infty\frac{n!m!}{(n+m+1)(n+m+1)!}\tag{4} \end{align} Now, consider \begin{align} a_n &=\sum_{m=0}^nm!(n-m)!\\ &=n!+\sum_{m=0}^{n-1}m!(n-m-1)!\,((n+1)-(m+1))\\ &=n!+(n+1)\sum_{m=0}^{n-1}m!(n-m-1)!-\sum_{m=0}^{n-1}(m+1)!(n-m-1)!\\ &=n!+(n+1)a_{n-1}-(a_n-n!)\tag{5}\\ a_n&=n!+\frac{n+1}{2}a_{n-1}\tag{6}\\ b_n&=\frac{2^n}{n+1}+b_{n-1}\quad\text{where }b_n=\frac{2^n}{(n+1)!}a_n\tag{7} \end{align} Thus $$a_n=\frac{(n+1)!}{2^n}\sum_{k=0}^n\frac{2^k}{k+1}\tag{8}$$ Using $(8)$, the sum in $(4)$ is \begin{align} \hspace{-1cm}\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{n!m!}{(n+m+1)(n+m+1)!} &=\sum_{m=0}^\infty\sum_{n=m}^\infty\frac{(n-m)!m!}{(n+1)(n+1)!}\tag{9}\\ &=\sum_{n=0}^\infty\sum_{m=0}^n\frac{(n-m)!m!}{(n+1)(n+1)!}\tag{10}\\ &=\sum_{n=0}^\infty\frac{a_n}{(n+1)(n+1)!}\tag{11}\\ &=\sum_{n=0}^\infty\sum_{k=0}^n\frac{2^{k-n}}{(n+1)(k+1)}\tag{12}\\ &=\sum_{k=0}^\infty\sum_{n=k}^\infty\frac{2^{k-n}}{(n+1)(k+1)}\tag{13}\\ &=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{2^{-n}}{(n+k+1)(k+1)}\tag{14}\\ &=\sum_{k=0}^\infty\frac1{(k+1)^2}+\sum_{k=0}^\infty\sum_{n=1}^\infty\frac{2^{-n}}{(n+k+1)(k+1)}\tag{15}\\ &=\frac{\pi^2}{6}+\sum_{n=1}^\infty\sum_{k=0}^\infty\frac{2^{-n}}{n}\left(\frac1{k+1}-\frac1{n+k+1}\right)\tag{16}\\ &=\frac{\pi^2}{6}+\sum_{n=1}^\infty\frac{H_n}{2^nn}\tag{17}\\ &=\frac{\pi^2}{4}\tag{18} \end{align} Explanation:
$\ \:(9)$: change variables $n\mapsto n-m$
$(10)$: change order of summation
$(11)$: use the definition of $a_n$ from $(5)$
$(12)$: apply $(8)$
$(13)$: change order of summation
$(14)$: change variables $n\mapsto n+k$
$(15)$: pull out the $n=0$ terms
$(16)$: evaluate $\zeta(2)$ and use partial fractions
$(17)$: $H_n=\sum\limits_{k=0}^\infty\left(\frac1{k+1}-\frac1{n+k+1}\right)$
$(18)$: apply sum $(8)$ from this answer

• GREAT ANSWER! (+1) Commented Dec 27, 2013 at 8:45