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Topic: EVALUATE THIS INFINITE SUM (Read 540 times) |
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pcbouhid
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EVALUATE THIS INFINITE SUM
« on: Nov 20th, 2005, 4:55am » |
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Evaluate this infinite sum:
k=inf 1
SUM -------------------
k=0 (3k+1)(3k+2)(3k+3)
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| « Last Edit: Nov 20th, 2005, 4:58am by pcbouhid » |
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Eigenray
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Re: EVALUATE THIS INFINITE SUM
« Reply #1 on: Nov 20th, 2005, 6:03am » |
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Fun.
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 = e2 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
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Barukh
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Re: EVALUATE THIS INFINITE SUM
« Reply #2 on: Nov 20th, 2005, 10:11am » |
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on Nov 20th, 2005, 6:03am, Eigenray wrote:
Fun2 ! 
Is it possible in a similar manner to evaluate infinite sums of n-degree polynomials?
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