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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)
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  • 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
>



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