sum done two ways gives different answers in Mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg106788] sum done two ways gives different answers in Mathematica
- From: Roger Bagula <roger.bagula at gmail.com>
- Date: Sat, 23 Jan 2010 07:35:11 -0500 (EST)
The sum:
Sum[1/(((-1)^(-I)*2)^n*(n + 1)), {n, 0, Infinity}]
gives:
-2*(-1)^(-I)*Log[1-(-1)^I/2]
N[%]=1.010902
ComplexExpand[1/(((-1)^(-I)*2)^n*(n + 1))]
gives
2^(-n)*Exp[-n*Pi]/(1+n)
with Infinite sum:
Sum[2^(-n)*Exp[-n*Pi]/(1+n), {n, 0, Infinity}]
-2*Exp[Pi]*Log[1-Exp[-Pi]/2]
N[%]=0.010902
I thought this result was both interesting and puzzling.
This idea comes from:
Solve[Exp[I*Pi] - Exp[I*Log[x]] == 0, x]
{{x -> (-1)^(-I))}}
I thought that it might give something related to Pi
in an Infinite sum.
--
Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
http://www.google.com/profiles/Roger.Bagula
alternative email: roger.bagula at gmail.com