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        riddles >> hard >> Transcendental sum?
        
(Message started by: Eigenray on Feb 10^th, 2006, 9:04pm)

Title: Transcendental sum?
Post by Eigenray on Feb 10^th, 2006, 9:04pm

Evaluate
[sum] n⁵ / (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 [link=http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1136518344]related thread[/link].

Title: Re: Transcendental sum?
Post by Eigenray on Dec 19^th, 2006, 11:36am

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/(k-1)! [sum]_n=1^oo n^k-1e^{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 n⁵e^{2pi imnz}
= C [sum]_n n⁵ e^{2pi inz}/(1-e^{2pi inz})
= -C [sum]_n n⁵/(1-e^{-2pi inz})
= -C A(-2iz),

where C = 2(-2pi i)⁶/5!, and

A(x) = [sum]_n=1^oo n⁵/(1-e^{pi n x})
= [ 2Zeta(6) - G(ix/2) ] / C.

Similarly define

B(x) = [sum]_n=1^oo n⁵/(1+e^{pi n x}),

so that

A(x) + B(x) = [sum] n⁵ 2/(1-e^{2n pi x}) = 2A(2x).

Now the sum we are interested in is

S = [sum]_{{n odd}} n⁵/(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+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}} n^k/(1+e^{n pi}) = (2ⁿ-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.)