[seqfan] Re: Mathematical symbol
Richard J. Mathar
mathar at mpia-hd.mpg.de
Thu Oct 19 15:18:55 CEST 2023
The divergent series sum_{k>=1} (-1)^(k+1)*binomial(2k,k)/(2k-1)
can be written as
sum_{k>=1} (-1)^(k+1)*Gamma(2k+1)/(2k-1)/Gamma^2(k+1)
...by the duplication formula fo the Gamma-function..
=
sum_{k>=1} (-1)^(k+1)*Gamma(k+1/2)Gamma(k+1)/sqrt(pi)/2^[1-(2k+1)]/(2k-1)/Gamma^2(k+1)
=
sum_{k>=1} (-1)^(k+1)*Gamma(k+1/2)Gamma(k+1)/sqrt(pi)/2^[-2k]/(2k-1)/Gamma(k+1)/k!
=
sum_{k>=1} (-1)^(k+1)*Gamma(k+1/2)Gamma(k+1)/Gamma(1/2)/2^[-2k]/(2k-1)/Gamma(k+1)/k!
=
- (1/2) sum_{k>=1} (-1)^k*Gamma(k+1/2)Gamma(k+1)/Gamma(1/2)/2^[-2k]/(k-1/2)/Gamma(k+1)/k!
=
- (1/2) sum_{k>=1} (-4)^k*Gamma(k+1/2)Gamma(k+1)/Gamma(1/2)/(k-1/2)/Gamma(k+1)/k!
=
- (1/2) sum_{k>=1} (-4)^k*Gamma(k+1/2)/Gamma(1/2)/(-1/2+k)/k!
=
- (1/2) sum_{k>=1} (-4)^k*Gamma(k+1/2)/Gamma(1/2)/(-1/2)*(-1/2)_k/(1/2)_k/k!
=
sum_{k>=1} (-4)^k*Gamma(k+1/2)/Gamma(1/2)*(-1/2)_k/(1/2)_k/k!
=
sum_{k>=1} (-4)^k*(1/2)_k*(-1/2)_k/(1/2)_k/k!
=
sum_{k>=1} (-4)^k*(-1/2)_k/k!
=
-1+ sum_{k>=0} (-4)^k*(-1/2)_k/k!
=
-1+ 1F0(-1/2;;-4)
-1+ 2.23606797..
where the notation (a)_n = Gamma(a+n)/Gamma(a) is Pochhammer's symbol
So the formula is the evaluation of a hypergeometric series by its
analytic continuation outside the usual region of convergence.
--
Richard Mathar
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