The closed form of this zeta-type function for all $N\ge 0\in\mathbb{Z}$

Question

In F.M.S. Lima's paper on Riemann zeta-type functions, he conjectures the following formula: $$\sum_{n=1}^\infty{\frac{\zeta(2n)}{2n(2n+1)\dots(2n+N)}\frac{1}{4^n}}=\frac{1}{2}\left[\frac{\ln{\pi}}{N!}-\frac{H_N}{N!}+\sum_{m=1}^{(N-1)/2}(-1)^{m+1}\frac{\zeta(2m+1)}{\pi^{2m}(N-2m)!}\right]$$ For $N$ being a positive odd integer.

Assuming this is true, I feel that this can be extended to any positive integer $N$.

All I am going on is the fact that for $N=2$ we have $$\sum_{n=1}^\infty{\frac{\zeta(2n)}{2n(2n+1)(2n+2)}\frac{1}{4^n}}=\frac{1}{2}\left[\frac{\ln{\pi}}{2!}-\frac{H_2}{2!}+\frac{7}{4\pi^2}\zeta(3)\right]$$ According to WolframAlpha.

But sums like the following don't make any sense $$\frac{7}{4\pi^2}\zeta(3)=\sum_{m=1}^{1/2}(-1)^{m+1}\frac{\zeta(2m+1)}{\pi^{2m}(2-2m)!}$$

Then what is the generalized definition for the zeta sum for any $N\ge 0\in\mathbb{Z}$?

Answer 1

Hint: I think the analogous conjecture to F.M.S. Lima's linked to above, but instead for even $N\ge 2$ is

$$\sum_{n=1}^\infty{\frac{\zeta(2n)}{2n(2n+1)\dots(2n+N)}\frac{1}{4^n}}=\frac{1}{2}\left[\frac{\ln{\pi}}{N!}-\frac{H_N}{N!}+\sum_{m=1}^{(N-2)/2}(-1)^{m+1}\frac{\zeta(2m+1)}{\pi^{2m}(N-2m)!}+(-1)^{(N/2)+1}\left(\frac{2^{N+1}-1}{2^{N+1}}\right){\frac{{2}\zeta(N+1)}{\pi^{N}}}\right]$$

When $N=0$ the sum equals $[ln({\pi/2})/2]$ and when $N=1$ the sum equals the first two terms of Lima's conjectural formula. These first two results are widely known and easily proved (e.g. see Boros and Moll, Irresistible Integrals)

Answer 2

Let me just tell you that my conjecture, mentioned above, has been proved true (by myself) in Theorem 2 of: Annali di Matematica Pura ed Applicata vol.194, p.1015 (2015). There in Theorem 3 of this paper, I prove a generalization of Theorem 2 which may be useful for answering the question by tyobrien.

Prof. Fabio M. S. Lima