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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 $\pi^2/6$.

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  • $\begingroup$ You can do two of the sums, one at a time. $\endgroup$ – b.gates.you.know.what Dec 23 '13 at 9:10
  • $\begingroup$ 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. $\endgroup$ – Andrew Dec 23 '13 at 20:49
  • $\begingroup$ @Chris'ssis please see the answer below. Numerical checking give near enough values. $\endgroup$ – Andrew Dec 24 '13 at 8:27
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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 $\frac12\psi ^{(1)}\left(\frac{1}{2}\right)=\frac{\pi ^2}{4}$.

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  • 1
    $\begingroup$ This seems too nice to be true ... $\endgroup$ – user 1357113 Dec 24 '13 at 12:54
  • $\begingroup$ FindSequenceFunction is just a guess, since it operates on a finite portion of the sequence; is that not correct? $\endgroup$ – robjohn Dec 27 '13 at 8:50
  • $\begingroup$ @robjohn yes. $~$ $\endgroup$ – Andrew Dec 27 '13 at 9:01
  • $\begingroup$ Nonetheless, nice answer! (+1) $\endgroup$ – robjohn Dec 27 '13 at 9:04
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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

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  • 1
    $\begingroup$ GREAT ANSWER! (+1) $\endgroup$ – user 1357113 Dec 27 '13 at 8:45

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