How would you tackle this series by real analysis? \sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m^4(m^2+n^2)}

Zaynah Dunn

Zaynah Dunn

Answered question

2022-02-22

How would you tackle this series by real analysis?
n=1m=11m4(m2+n2)

Answer & Explanation

Jett Brooks

Jett Brooks

Beginner2022-02-23Added 7 answers

Since:
n11m2+n2=1+mπcoth(mπ)2m2
our sum equals:
ζ(6)2+π2m1coth(mπ)m5=πζ(5)ζ(6)2+πm11m5(e2mπ1) (1)
and maybe the last series has a nice closed form.
Then:
m11m5(e2mπ1)=m1Li5(e2πm)=12πicic+iΓ(z)(2π)zζ(z+5)ζ(z)dz
If now we set f(z)=Γ(z)(2π)zζ(z+5)ζ(z), we have:
Res(f(z), z=0)=ζ(5)2
Res(f(z), z=1)=π5540
Res(f(z), z=3)=π5540
Res(f(z), z=4)=2π43ζ(4)
Res(f(z), z=5)=π51890
hence a closed form for (1) just depends on a closed form for ζ(4), or, by the reflection formula, on a closed form for:
ζ(5)=n1lognn5

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