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   Transcendental sum?
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   Author  Topic: Transcendental sum?  (Read 1029 times)
Eigenray
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Transcendental sum?  
« on: Feb 10th, 2006, 9:04pm »
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Evaluate
[sum] n5 / (1 + en pi),
where the sum is over positive odd n=1,3,5,....  Generalize.
 
Since this is hard, I'll provide a hint right from the start: consider
G(z) = [sum] (n + mz)-6,
where the sum is over all pairs of integers (n,m) not both zero.  Complex numbers are involved, but no complex analysis is necessary if you're willing to accept a result from a related thread.
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Eigenray
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Re: Transcendental sum?  
« Reply #1 on: Dec 19th, 2006, 11:36am »
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Okay, I guess this one was hard, but the answer is pretty amazing, so I'll give it.  We have
 
G(z) = [sum] (n+mz)-6
 = [sum]n n-6 + 2[sum]m=1oo [sum]n (n+mz)-6,
 
where the first sum counts pairs with m=0, and is just 2Zeta(6), and the other sum takes m and -m together, for m>0.  From the other thread, we have
 
[sum] (n+t)-k = (-2pi i)k/(k-1)! [sum]n=1oo nk-1e2pi i nt,
 
and so, changing the order of summation to sum the geometric series, we get
 
G(z) - 2Zeta(6) = 2(-2pi i)6/5! [sum]m,n>0 n5e2pi imnz
 = C [sum]n n5 e2pi inz/(1-e2pi inz)
 = -C [sum]n n5/(1-e-2pi inz)
 = -C A(-2iz),
 
where C = 2(-2pi i)6/5!, and
 
A(x) = [sum]n=1oo n5/(1-epi n x)
 = [ 2Zeta(6) - G(ix/2) ] / C.
 
Similarly define
 
B(x) = [sum]n=1oo n5/(1+epi n x),
 
so that
 
A(x) + B(x) = [sum] n5 2/(1-e2n pi x) = 2A(2x).
 
Now the sum we are interested in is
 
S = [sum]{n odd} n5/(1+en pi)
 = B(1) - 32B(2)
 = [2A(2) - A(1)] - 32[2A(4) - A(2)]
 = -A(1) + 34A(2) - 64A(4)
 = 2Zeta(6)/C [-1+34-64] - 1/C [-G(i/2) + 34G(i) - 64G(2i)].
 
But observe that
 
G(-1/z) = [sum] (n-m/z)-6 = z6 [sum] (-m + nz)-6 = z6 G(z),
 
which gives
 
G(i) = G(-1/i) = i6 G(i) = -G(i),
 
so G(i)=0, and
 
G(i/2) = G(-1/(2i)) = (2i)6 G(2i) = -64 G(2i).
 
Thus in fact
 
S = -31*2Zeta(6)/C
 = -31[1/42 26 pi6/6!]/[2(-2pi i)6/5!]
 = 31/504.
 
More generally, if k=4r+1, r>0, then
 
[sum]{n odd} nk/(1+en pi) = (2n-1)Bn+1/[2(n+1)],
 
where Bn is Bernoulli.  In particular, this is rational!
 
(This seems to hold for k=1, too, but the proof would require more care, since then G is not absolutely convergent.)
« Last Edit: Dec 19th, 2006, 11:40am by Eigenray » IP Logged
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