This time I remembered to include the exponent $k$!
As mentioned in a previous post, this answer to this is key to finding an explicit expression for the sums of the powers of the reciprocals of the figurate numbers, for powers greater then or equal to $2$.
Aside from the beautiful patterns that may result, this approach may provide new relationships among the zeta and other functions. For example, consider triangular numbers ($a=1$) where $k=8$. The expression is:
$\begin{align*}1/1^8 &+ 1/3^8 + \dots =\\ &= 2^9\left[\binom70\zeta(8) + \binom92\zeta(6) + \binom{11}4\zeta(4) + \binom{13}6\zeta(2) + \binom{15}7\zeta(0)\right] \end{align*}$
where $\zeta(x)$ is the Riemann zeta function with $\zeta(0)=-1/2$.
It's apparent that for an exponent $k$, I can find a well-defined expressions in the form of $c(i)\zeta(2i)$, where $c(i)$ is a coefficient like $2^9\binom70$.
So, in the limit of $k=\infty$, we have something like $\sum_{i=1}^\infty c(i)\zeta(2i)=1$, giving us a possibly new relationship among the even-valued zeta.