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I need a closed form for the following series which appeared when evaluating another sum. $$S(m,x)=\sum_{n=1}^\infty\binom{2n}{n}\zeta(m,n+1)x^n, \quad m\in\mathbb{Z}$$

I've tried expanding the Hurwitz zeta function by $(25.11.10)$, $$\zeta(m,n+1)=\sum_{k=0}^\infty\frac{(m)_{k}}{k!}\zeta(m+k)(-n)^k$$ and switching the order of summation, but the result doesn't look any better.

Any ideas on how to find the generating function?

bob
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1 Answers1

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From here we have a identity:

$$\sum _{n=1}^{\infty } \zeta (m,n+1) x^n=\frac{x \zeta (m)-\text{Li}_m(x)}{1-x}$$ and using the answer from here we have:

for any $|x|<\frac{1}{4}$

$$S(m,x)=\\\sum _{n=1}^{\infty } \zeta (m,n+1) \binom{2 n}{n} x^n=\\\int_0^{\frac{\pi }{2}} \left(-\frac{2 \text{Li}_m\left(4 x \cos ^2(t)\right)}{\pi \left(1-4 x \cos ^2(t)\right)}+\frac{8 x \cos ^2(t) \zeta (m)}{\pi \left(1-4 x \cos ^2(t)\right)}\right) \, dt=\\\left(\frac{1}{\sqrt{1-4 x}}-1\right) \zeta (m)-\int_0^{\frac{\pi }{2}} \frac{2 \text{Li}_m\left(4 x \cos ^2(t)\right)}{\pi \left(1-4 x \cos ^2(t)\right)} \, dt$$

for the last integral Mathematica can compute for $m=0$ and $m=1$.

Mariusz Iwaniuk
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  • You're overcomplicating things. Your first term should be $$\zeta(m)\left[\frac1{\sqrt{1-4x}}-1\right]$$ To see this, just write $\zeta(m,n+1)=\zeta(m)-H^{(m)}n$, and split the series. $\sum{n=1}^\infty\zeta(m)\binom{2n}{n}x^n$ should be your first term (equal to the expression I just gave). Your second term should be the one discussed in the linked answer (equal to that integral). – Christian E. Ramirez Aug 19 '23 at 10:54