Somewhat simplified, Sum[1/Fibonacci[n], {n,∞}] == Sqrt[5]/(4*Log[ϕ])* (2*Im[QPolyGamma[0, 1/2 + I*Pi/(4*Log[ϕ]), 1/ϕ^2]] - QPolyGamma[0, 1, ϕ^2] + Re[QPolyGamma[0, 1 - (I*Pi)/(2*Log[ϕ]), ϕ^2]]) where ϕ = ϕ^2-1 ~ 1.618 . We could call QPolyGamma[0,...] QDigamma: Subscript[\[Psi], q][z] == D[Subscript[\[CapitalGamma], q][z], z]/Subscript[\[CapitalGamma], q][z] == -Log[1 - q] + Log[q]*Sum[q^(n + z)/(1 - q^(n + z)), {n, 0, Infinity}] On Wed, May 21, 2014 at 11:54 AM, Bill Gosper <billgosper@gmail.com> wrote:
David Wilson>If you considered
f(n, x) = SUM(n = 0..inf, 1 / (x^n + 1))
I think you mean f(1,x).
to be a closed form, could we find a closed form for
f(a, b) = SUM(n = 0..inf, 1 / (a^n + b^n))?
Mathematica doesn't think so, even with its all-purpose In[86]:= Sum[1/(1 - a*x^n), n]
Out[86]= -((-n Log[x] + QPolyGamma[0, n - Log[1/a]/Log[x], x])/Log[x])
(Note indefinite sum.) Which can even, with some work, express the ever-nagging
Out[139]= (1/(4 ArcCsch[2])) Sqrt[5] (-QPolyGamma[0, 1, GoldenRatio^2] + I (QPolyGamma[0, 1/2 - (I π)/(4 ArcCsch[2]), 1/GoldenRatio^2] -
QPolyGamma[0, 1/2 + (I π)/(4 ArcCsch[2]), 1/GoldenRatio^2]) + Re[QPolyGamma[0, 1 - (I π)/(2 ArcCsch[2]), GoldenRatio^2]])
In[140]:= N[{%, Sum[1/Fibonacci[n], {n, ∞}]}]
Out[140]= {3.35989 + 0. I, 3.35989}
--rwg