7
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This sum

s = Sum[Gamma[k/2]/(2 k!), {k, 1, ∞}]

$\sum _{k=1}^{\infty } \frac{\Gamma \left(\frac{k}{2}\right)}{2 k!}$

is returned unevaluated (version 10.1.0).

However, if we split even from odd k the sums are done

se = Sum[Gamma[k]/(2 (2 k)!), {k, 1, ∞}]

(*
  1/4 HypergeometricPFQ[{1, 1}, {3/2, 2}, 1/4]
*)

so = Sum[Gamma[k - 1/2]/(2 (2 k - 1)!), {k, 1, ∞}]

(*
  1/2 π Erfi[1/2]
*)

And the sum $se + so$ is indeed $s$, as can be checked numerically.

se + so // N

(* 1.23826 *)

sn = NSum[Gamma[k/2]/(2 k!), {k, 1, ∞}]

(* 1.23826 *)

Question: how can I get Mathematica to discover this simple splitting by itself?

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  • $\begingroup$ Mathematica is not an AI program; at least, not yet. So sometimes you have to do the thinking while Mathematica does the drudge work. $\endgroup$ – m_goldberg Nov 20 '15 at 9:46
  • $\begingroup$ @m_goldberg You might know me as a strong advocat of "man-machine-interaction" in solving MMA problems, and have shown this on several occasions here, including this one. $\endgroup$ – Dr. Wolfgang Hintze Nov 20 '15 at 11:01
  • $\begingroup$ But I still don't understand what you mean by "tell" and "discover for itself" if not AI, or what kind of an answer you expect $\endgroup$ – m_goldberg Nov 20 '15 at 11:18
  • $\begingroup$ @m_goldberg I didn't use the word "tell" as you suggest. Let me summarize: I have shown how to derive the exact symbolic result of the sum. My question is just if there is perhaps some command like Simplify etc. which helps MMA to find the result. BTW this is the first instance in more than 11 years of using MMA that someone comes up with the term AI in such a simple context ;-) $\endgroup$ – Dr. Wolfgang Hintze Nov 20 '15 at 12:04
  • $\begingroup$ A workaround: Sum[Gamma[k/2]/(2 k Gamma[k]), {k, 1, ∞}] $\endgroup$ – J. M. is away Nov 20 '15 at 14:08
4
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Just regularize:

 Sum[Gamma[k/2]/(2 k!), {k, 1, ∞}, Regularization -> "Abel"]

1/4 (2 π Erfi[1/2] + HypergeometricPFQ[{1, 1}, {3/2, 2}, 1/4])

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  • $\begingroup$ Thank you very much @yarchik that's what I was looking for. You can also use the neutral form Regularization->True $\endgroup$ – Dr. Wolfgang Hintze Nov 20 '15 at 18:54
0
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Maybe this give you some idea:

    Plus[Sequence @@ Table[Gamma[k/2]/k!, {k, 1, 1000}]] // N
so = Plus[
   Sequence @@ Table[Gamma[k - 1/2]/(2 (2 k - 1)!), {k, 1, 1000}]] // N
se = Plus[Sequence @@ Table[Gamma[k]/(2 k)!, {k, 1, 1000}]] // N
se + so

I know this is not an answer but I am under 50 reputation and can not post comment.

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  • $\begingroup$ Thanks for your answer. But you will find that I did just this - and complete up to k = oo - in my question. $\endgroup$ – Dr. Wolfgang Hintze Nov 20 '15 at 12:39

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