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Say you had $$\sum^{1287}_{n=1} (n!)^4$$ How would you got about summing this as it is not arithmetic you cannot use the formula. I tried creating a common difference in terms of $n$ but that did not work for me.

Tiernan Watson
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    Are you aware that the result will be a 3445 digit number? – MJD Dec 10 '15 at 17:17
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    What makes you think there is an simpler answer? – fleablood Dec 10 '15 at 17:20
  • If you are looking for a general closed form for these kind of sums, you are probably doomed to fail. Even the sum of factorials (not raised to any power) is a complicated sum to calculate. E.g. see http://math.stackexchange.com/questions/227551/sum-k-1-2-3-cdots-n-is-there-a-generic-formula-for-this – Colm Bhandal Dec 10 '15 at 17:20
  • Im not getting which type of logic you have to this sum as $1287!=\infty$ cant be calculated – Archis Welankar Dec 10 '15 at 17:24
  • @ArchisWelankar what are you talking about, 1287! is not infinite, it's just a number. – Colm Bhandal Dec 10 '15 at 17:25
  • Ya im just saying that for us to calculate $1287!$ its next to impossible we would need a computer or some software to calculate it. – Archis Welankar Dec 10 '15 at 17:33
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    @ArchisWelankar OK, but why not just say that? As mathematicians we should be aspiring towards the dizzying heights of universes so large they blow our minds. This number is petty compared to, e.g. Graham's number: https://www.youtube.com/watch?v=GuigptwlVHo. Which is still just a number, and so compared to infinity it is petty. And infinity itself has different flavours, as Cantor showed. All of this, beyond what's merely humanly calculable. – Colm Bhandal Dec 10 '15 at 17:39

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It seems you're looking for a closed form solution. I doubt this exists. One possible rearrangement of the terms, which may be faster to compute with a computer program, uses nesting:

$$\sum^{n}_{k=1} (k!)^a = 1 + 2^a(1 + 3^a (1 + 4^a (1 + \dots + (n-1)^a(1 + n^a)\dots))))$$

Written using recursive notation, that is:

$$f_{a,n}(k)=\left\{\begin{array}{cc} k^a(1 + f_{a,n}(k+1)) & k \leq n \\ 0 & k>n \end{array}\right.$$

We then simply calculate $f_{a,n}(1)$ to get the above. In your case that would be $f_{4,1287}(1)$.

Colm Bhandal
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