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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.