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Topic: Transcendental sum? (Read 1029 times) 

Eigenray
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Transcendental sum?
« on: Feb 10^{th}, 2006, 9:04pm » 
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Evaluate [sum] n^{5} / (1 + e^{n 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 19^{th}, 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=1}^{oo} [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}/(k1)! [sum]_{n=1}^{oo} n^{k1}e^{2pi 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} n^{5}e^{2pi imnz} = C [sum]_{n} n^{5} e^{2pi inz}/(1e^{2pi inz}) = C [sum]_{n} n^{5}/(1e^{2pi inz}) = C A(2iz), where C = 2(2pi i)^{6}/5!, and A(x) = [sum]_{n=1}^{oo} n^{5}/(1e^{pi n x}) = [ 2Zeta(6)  G(ix/2) ] / C. Similarly define B(x) = [sum]_{n=1}^{oo} n^{5}/(1+e^{pi n x}), so that A(x) + B(x) = [sum] n^{5} 2/(1e^{2n pi x}) = 2A(2x). Now the sum we are interested in is S = [sum]_{{n odd}} n^{5}/(1+e^{n pi}) = B(1)  32B(2) = [2A(2)  A(1)]  32[2A(4)  A(2)] = A(1) + 34A(2)  64A(4) = 2Zeta(6)/C [1+3464]  1/C [G(i/2) + 34G(i)  64G(2i)]. But observe that G(1/z) = [sum] (nm/z)^{6} = z^{6} [sum] (m + nz)^{6} = z^{6} G(z), which gives G(i) = G(1/i) = i^{6} 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 2^{6} pi^{6}/6!]/[2(2pi i)^{6}/5!] = 31/504. More generally, if k=4r+1, r>0, then [sum]_{{n odd}} n^{k}/(1+e^{n pi}) = (2^{n}1)B_{n+1}/[2(n+1)], where B_{n} 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 19^{th}, 2006, 11:40am by Eigenray » 
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