Re: Bug in infinite sum

*To*: mathgroup at smc.vnet.net*Subject*: [mg127396] Re: Bug in infinite sum*From*: Bob Hanlon <hanlonr357 at gmail.com>*Date*: Sun, 22 Jul 2012 04:30:25 -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*: <20120721034220.50F3968C9@smc.vnet.net>

Must be a problem with the version that you are running. $Version "8.0 for Mac OS X x86 (64-bit) (October 5, 2011)" s[x_] = Sum[Binomial[x, k], {k, 0, Infinity}] 2^x u[x_] = Sum[Binomial[x, 2*k + 1], {k, 0, Infinity}] 2^(-1 + x) g[x_] = Sum[Binomial[x, 2*k], {k, 0, Infinity}] 2^(-1 + x) s[x] == u[x] + g[x] True Bob Hanlon On Fri, Jul 20, 2012 at 11:42 PM, Dr. Wolfgang Hintze <weh at snafu.de> wrote: > 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 >

**References**:**Bug in infinite sum***From:*"Dr. Wolfgang Hintze" <weh@snafu.de>