0
$\begingroup$

The expression I'm trying to do something with is

Sum[1/(a!(s-a)!)b^(-(2a+1)/2) Sqrt[\[Pi]](2a)!/(4^a (a)!)Pochhammer[-a,k],{a,k,s}] where s and k are positive integers.

Mathematica is telling me it's ComplexInfinity since it uses some kind of more general functions which have poles for integer values. I tried using limits:

Limit[Sum[1/(a!(s-a)!)b^(-(2a+1)/2) Sqrt[\[Pi]](2a)!/(4^a (a)!)Pochhammer[-a,k],{a,k-x,s}],x->0]

And it gives some kind of expression but it spits complex infinities here and there and is very slow to work with since you need to introduce more limits for integer k and s.

Is there a proper way to simplify the sum into something?

$\endgroup$
  • $\begingroup$ Using MMA 12.0: Sum[1/(a! (s - a)!) b^(-(2 a + 1)/ 2) Sqrt[\[Pi]] (2 a)!/(4^a (a)!) Pochhammer[-a, k] // FullSimplify, {a, k, s}] return imput. That means he doesn't know the answer. $\endgroup$ – Mariusz Iwaniuk Aug 23 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.