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        riddles >> hard >> EVALUATE THIS INFINITE SUM
        
(Message started by: pcbouhid on Nov 20^th, 2005, 4:55am)

Title: EVALUATE THIS INFINITE SUM
Post by pcbouhid on Nov 20^th, 2005, 4:55am

Evaluate this infinite sum:

k=inf 1
SUM -------------------
k=0 (3k+1)(3k+2)(3k+3)

No computers allowed!

Title: Re: EVALUATE THIS INFINITE SUM
Post by Eigenray on Nov 20^th, 2005, 6:03am

Fun.
[hide]First rewrite it as
2S = [sum] 1/(3k+1) - 2/(3k+2) + 1/(3k+3)
Now let
L(x) = x + x²/2 + x³/3 + ... = -log(1-x).
Ignoring convergence,
2(w-w')S = L(w) - L(w') - w'L(w) + wL(w')
= (L(w)+L(w'))*(w-w')/2 + (L(w)-L(w'))*3/2,
where w = e^{2 pi i/3}, w'=w² are cube roots of 1. Evaluating
L(w)+L(w') = -log[(1-w)(1-w')] = -log|1-w|² = -log 3,
L(w)-L(w') = log[(1-w')/(1-w)] = log (-w') = i pi/3,
it follows
2i sqrt(3) S = -(log 3) i sqrt(3)/2 + i pi/2,
so S = [ pi/sqrt(3) - log 3 ]/4[/hide]

Title: Re: EVALUATE THIS INFINITE SUM
Post by Barukh on Nov 20^th, 2005, 10:11am

on 11/20/05 at 06:03:43, Eigenray wrote:

Fun.

Fun² ! :D

Is it possible in a similar manner to evaluate infinite sums of n-degree polynomials?