# A triple sum related question

I'm trying to compute the triple sum

Sum[ 1/(i! j! k! ), {i, 1, Infinity}, {j, i + 1, Infinity}, {k, j + 1, Infinity}]


but Mathematica doesn't return any value.

What else can I do here? Thanks!

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Maybe because it does not have an exact values in terms of $\pi$, $e$ and other known constants or functions. N[Sum[1/(i! j! k!), {i, 1, Infinity}, {j, i + 1, Infinity}, {k, j + 1, Infinity}]] computes fine and returns 0.122759. – m0nhawk Jan 21 '13 at 13:18
Maybe NSum is sufficient in your case ? – b.gatessucks Jan 21 '13 at 13:20
@m0nhawk: I'm sure that in the answer must also be the constant $e$. The triple sums also involves the hypergeometric function but this shouldn't be a problem, right? – Chris's sister Jan 21 '13 at 13:21
@Chris'ssister: 1. It was just an assumption why it does not compute nothing. 2. That a function that gives the numerical value of expression, see N. – m0nhawk Jan 21 '13 at 13:24
@m0nhawk: thank you very much for the explanations offered! – Chris's sister Jan 21 '13 at 13:26

One can get an explicit result using symmetric functions. First rewrite the infinite sum as a finite sum:

$$S_n=\sum_{i=1}^{n-2}\sum_{j=i+1}^{n-1}\sum_{k=j+1}^{n}{1\over{i!j!k!}}$$

Now define the power sums:

$$s_j=\sum_{k=1}^{n} (k!)^{-j}$$

We can now rewrite $S_n$ as a function of $s_1$, $s_2$, and $s_3$:

$$S_n=(2 s_3 - 3 s_2 s_1 + s_1^3)/6$$

The limit can now be found by finding the limits of the power sums and then plugging in those limits to the previous equation:

s1Limit = Sum[1/i!, {i, 1, ∞}]
(* -1 + e *)
s2Limit = Sum[1/(i!)^2, {i, 1, ∞}]
(* -1  +BesselI[0,2] *)
s3Limit = Sum[1/(i!)^3, {i, 1, ∞}]
(* -1 + HypergeometricPFQ[{},{1,1},1] *)
sLimit = (2 s3Limit - 3 s2Limit s1Limit + s1Limit^3)/6


A numeric approximation is

N[sLimit]
(* 0.12275911513957549 *)

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