Re: Bug in infinite sum

*To*: mathgroup at smc.vnet.net*Subject*: [mg127403] Re: Bug in infinite sum*From*: Peter Pein <petsie at dordos.net>*Date*: Sun, 22 Jul 2012 04:32:46 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: mathgroup-newout@smc.vnet.net*Delivered-to*: mathgroup-newsend@smc.vnet.net*References*: <jud8jf$j2m$1@smc.vnet.net>

Am 21.07.2012 05:44, schrieb Dr. Wolfgang Hintze: > Consider this nice sum > > s[x_] = Sum[Binomial[x, k], {k, 0, Infinity}] > 2^x > > and that, (with the odd terms only) > > u[x_] = Sum[Binomial[x, 2*k + 1], {k, 0, Infinity}] > 2^(-1 + x) > > But now the even terms only ... and the surprise: > g[x_] = Sum[Binomial[x, 2*k], {k, 0, Infinity}] > > (Sqrt[Pi]*Gamma[1 + x]*GegenbauerC[x, 1/2 - x, 1])/ > (2^x*Gamma[1/2 + x]) > > looks complicated, but let's see > FullSimplify[g[x], x > 0] > > 2^x*Cos[Pi*x] > > much simpler, but definitely wrong (giving e.g. 0 for x=1/2) > Of course g should be s - u = 2^x -1/2 2^s = 1/2 2^s. > > Best regards, > Wolfgang > Hmm on version 8 (Windows 64-bit) I get: Sum[Binomial[x, 2*k], {k, 0, Infinity}] 2^(-1+x) and the same result for Sum[Binomial[x, k], {k, 0, Infinity, 2}] as it should be. Peter