Mathematica Stack Exchange is a question and answer site for users of Mathematica. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|improve this question
2  
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

Result

A numeric approximation is

N[sLimit]
(* 0.12275911513957549 *)
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.