       Re: sum done two ways gives different answers in

• To: mathgroup at smc.vnet.net
• Subject: [mg106829] Re: [mg106788] sum done two ways gives different answers in
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Sun, 24 Jan 2010 05:43:08 -0500 (EST)

```I get the same result both ways.

\$Version

7.0 for Mac OS X x86 (64-bit) (February 19, 2009)

expr1 = Sum[1/(((-1)^(-I)*2)^n*(n + 1)),
{n, 0, Infinity}] // Simplify

(-2*Log[1 - (-1)^I/2])/(-1)^I

expr1 // N

1.01096

expr2 = ComplexExpand[
1/(((-1)^(-I)*2)^n*(n + 1))] //
Simplify // Quiet

E^(Pi*(-n))/(2^n*(n + 1))

expr3 = Sum[expr2,
{n, 0, Infinity}] // Simplify

-2*E^Pi*Log[1 - 1/(E^Pi*2)]

expr1 == expr3 // FullSimplify

True

N[expr3]

1.01096

Bob Hanlon

---- Roger Bagula <roger.bagula at gmail.com> wrote:

=============
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 :