Series $\sum _{n=0}^{\infty } \frac{1}{(2 n+1)^m (2 n+3)^m (2 n+5)^m }$ and zeta function

Question

Consider the following series:

$$I_m=\sum _{n=0}^{\infty } \frac{1}{(2 n+1)^m (2 n+3)^m (2 n+5)^m } $$

Using Mathematica, it turns out that:

$$I_{2m}= P_{m0}-\sum_{k=1}^m P_{mk} \zeta(2k)$$

While:

$$I_{2m+1}= Q_{m0}-\sum_{k=1}^m Q_{mk} \zeta(2k+1)$$

Where $P_{mk}, Q_{mk}$ are rational numbers.

The rational numbers' denominators and numerators get huge for larger $m$, which is why I wanted to ask if we can derive a general closed form for $Q_{mk}$?

I'm less interested in the even case, since even zetas have a closed form in terms of powers of $\pi$.

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First few cases are:

$$I_1= \frac{1}{12} \\ I_3 = \frac{698-567 \zeta (3)}{55296}$$

$$I_5=\frac{868058-331695 \zeta (3)-451980 \zeta (5)}{509607936}$$

$$I_7=\frac{507671866-123130287 \zeta (3)-210713076 \zeta (5)-139985496 \zeta (7)}{2348273369088}$$

Answer 1

The question essentially boils down to finding partial fractions of

$$\frac1{(2n+1)^m(2n+3)^m(2n+5)^m}=\sum_{k=1}^m\frac{a_k}{(2n+1)^k}+\frac{b_k}{(2n+3)^k}+\frac{c_k}{(2n+5)^k}$$

which can be done by multiplying both sides by $(2n+a)^m$ and computing the terms via Taylor's theorem:

$$\frac1{(2n+3)^m(2n+5)^m}=\sum_{k=1}^ma_k(2n+1)^{m-k}+\frac{b_k(2m+1)^m}{(2n+3)^k}+\frac{c_k(2m+1)^m}{(2n+5)^k}$$

$$a_k=\frac{2^{k-3m}}{(m-k)!}\frac{\mathrm d^{m-k}}{\mathrm dx^{m-k}}\frac1{(x+\frac32)^m(x+\frac52)^m}\bigg|_{x=-1/2}=\frac1{(m-k)!}\sum_{j=0}^{m-k}d_j(-1)^{m-k}2^{k-4m-j}$$

$$b_k=\frac{2^{k-3m}}{(m-k)!}\frac{\mathrm d^{m-k}}{\mathrm dx^{m-k}}\frac1{(x+\frac12)^m(x+\frac52)^m}\bigg|_{x=-3/2}=\frac1{(m-k)!}\sum_{j=0}^{m-k}d_j(-1)^{m-j}2^{k-3m}$$

$$c_k=\frac{2^{k-3m}}{(m-k)!}\frac{\mathrm d^{m-k}}{\mathrm dx^{m-k}}\frac1{(x+\frac12)^m(x+\frac32)^m}\bigg|_{x=-5/2}=\frac1{(m-k)!}\sum_{j=0}^{m-k}d_j2^{k-4m-j}$$

where

$$d_j=\binom{m-k}j\frac{(2m-k-j-1)!(m+j-1)!}{(m-1)!}$$

(hopefully I did that right >.<)

The problem thus remains to compute $\sum_{n=0}^\infty\frac1{(2n+1)^m}$, which it is easy to see that

$$\sum_{n=0}^\infty\frac1{(2n+1)^m}=(1-2^{-m})\zeta(m)$$

and thus we have

\begin{align}I_m&=e_m+\sum_{k=2}^m\frac{(1-2^{-k})\zeta(k)}{(m-k)!}\sum_{j=0}^{m-k}d_j[(-1)^{m-k}2^{k-4m-j}+(-1)^{m-j}2^{k-3m}+2^{k-4m-j}]\end{align}

where $e_m$ is given by the leftovers from reindexing the sums for the $(2n+3)$ and $(2n+5)$ as well as the $k=1$ case.