1
$\begingroup$

I would like to have the following sequence, $(a)_n$, in a series where $a_0 = 1$ and $a_n = a(a+1)\cdots (a+ n -1)$. The series I have is $$ 2\sum_{n=0}^{\infty}\frac{(1/2)_n}{n!(2n+1)^3} $$ I know how to represent series in mathematica but I have no idea how I can insert the sequence into the series code. What I have is

2 Sum[Sequence Here/(n!(2n+1)^3),{n,0,infinity}]

Is it possible to do this in Mathematica?

$\endgroup$
  • $\begingroup$ Are you looking for Pochhammer perhaps? $\endgroup$ – ciao Feb 21 '15 at 1:52
  • $\begingroup$ @rasher yes, that worked. Should I delete or do you want provide an answer? $\endgroup$ – dustin Feb 21 '15 at 1:54
  • $\begingroup$ Sure, I'll answer... future reference. $\endgroup$ – ciao Feb 21 '15 at 1:56
1
$\begingroup$

You're after Pochhammer:

Pochhammer[a, 5]

(* a (1 + a) (2 + a) (3 + a) (4 + a) *)

Do note, the Mathematica definition by default is opposite that used in many fields: it is the Rising Factorial, while the traditional form can be confused with Falling Factorial, the perhaps more common use of Pochhammer...

$\endgroup$
  • $\begingroup$ @2012rcampion Oops... $\endgroup$ – ciao Feb 21 '15 at 2:38
1
$\begingroup$

You can also construct that series yourself, should you need that in another context in the future:

sa[n_]:=Product[a+i,{i,0,n-1}];

and insert that into your formula:

2 Sum[sa[n]/(n!(2 n+1)^3), {n,0,\[Infinity]}]

or avoid the function and write:

2 Sum[Product[a+i,{i,0,n-1}]/(n!(2 n+1)^3), {n,0,\[Infinity]}]

both giving you the result:

(1/(16 Gamma[3/2-a]))Sqrt[\[Pi]]
  Gamma[1-a] (\[Pi]^2+2 PolyGamma[0,1/2]^2 - 
   4 PolyGamma[0,1/2] PolyGamma[0,3/2-a] + 
   2 PolyGamma[0,3/2-a]^2 - 2 PolyGamma[1,3/2-a])
$\endgroup$

Your Answer

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

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