# Evaluate an infinite series involving the polygamma function OR first derivative of the hurwitz zeta function

Can we find a closed form for $$\sum _{k=1}^\infty\frac{\left(-1\right)^k}{2k-1}\left(2k^2-k+8k^2P_1(k)-16kP_2(k)+16P_3(k)\right)$$ where $$P_n(k)=\psi^{(-n)}\left(k+\frac12\right)-\psi^{(-n)}\left(k+1\right)$$

Here the definition for $\psi^{(-n)}(x)$ I'm using is $$\psi^{(-n)}(x)=\frac{1}{\left(n-2\right)!}\int _0^x\left(x-t\right)^{n-2}\ln\Gamma(t)dt$$ for $n\gt1$ and $\psi^{(-1)}(x)=\ln\Gamma(x)$

The value of the sum is near $5.7971...$

I've tried to simplify the $P_n(k)$ hoping that there would be cancelling, but with no luck. Perhaps there is an asymptotic approach?

EDIT: I've simplified the sum to be in terms of the first derivative of the Hurwitz Zeta function. So my sum is the same as

$$\frac{\pi}{4}(8\ln A+3\ln(2\pi))+\sum _{k=1}^\infty\frac{\left(-1\right)^k}{2k-1}\left[8k^2\left(\zeta^{(1,0)}\left(0,k+\frac12\right)-\zeta^{(1,0)}\left(0,k+1\right)\right)-16k\left(\zeta^{(1,0)}\left(-1,k+\frac12\right)-\zeta^{(1,0)}\left(-1,k+1\right)\right)+8\left(\zeta^{(1,0)}\left(-2,k+\frac12\right)-\zeta^{(1,0)}\left(-2,k+1\right)\right)\right]$$ Where $A$ is the Glaisher-Kinkelin constant. Maybe we can try using Perron's Formula?

• At $n=1$, the integral diverges. – Simply Beautiful Art Aug 27 '17 at 16:09
• Personally, I'd have$$\psi^{(-n)}(x)=\frac1{(n-1)!}\int_1^x(x-t)^{n-1}\psi(t)~\mathrm dt$$ – Simply Beautiful Art Aug 27 '17 at 16:12
• How'd you get $2.89855$? – Simply Beautiful Art Aug 27 '17 at 16:15
• Shouldn't that integral be from 0 to x? – tyobrien Aug 27 '17 at 16:23
• – Simply Beautiful Art Aug 27 '17 at 16:23

By Euler summing up to $0\le k\le p\le n\le15$, I found that

n    sum
---  ------------
0    3.6963863814
1    4.9242922276
2    5.4146695729
3    5.6246336957
4    5.7178928561
5    5.7602644696
6    5.7798140328
7    5.7889347699
8    5.7932259771
9    5.7952583022
10   5.7962259359
11   5.7966886603
12   5.7969107445
13   5.7970176641
14   5.7970692763
15   5.7970942481


Where I used

$$S_n=\sum_{p=0}^n\frac1{2^{p+1}}\sum_{k=0}^p\binom pka_{k-1}$$

where $a_k$ are your terms.