For $s=0,1,\dots$ and $n=1,2,\dots$, denote $r_{n,s}=\sum_{k=1}^n k^s$. It is well-known that $r_{n,s}$ are polynomials in $n$ with leading term $\frac{1}{s+1}n^{s+1}$. Let $R_{n,s}$ be the $(s+1)\times (s+1)$ (symmetric) matrix $$R_{n,s}:=\left( r_{n,i+j-2} \right)_{i=1,\dots,s+1}^{j=1,\dots,s+1}.$$
Let $H_{n,s}$ be the ``leading term'' of $R_{n,s}$, namely $$ H_{n,s}=\left( \frac{n^{i+j-1}}{i+j-1} \right)_{i,j}$$ and $R_{n,s}=H_{n,s}+O(n^{-1}) H_{n,s}$.
It is an experimental "fact" that $R_{n,s}$ is invertible, with entries rational in $n$, whose "leading term" is given by the matrix $$ T_{n,s}:=H_{n,s}^{-1}= \left( \frac{q_{i,j}}{n^{i+j-1}} \right)_{i,j} $$ where $q_{i,j}$ are the entries of the corresponding inverse Hilbert matrix, i.e. if $\cal{H}^{(s)}=\left( \frac{1}{i+j-1} \right)_{i,j}$ then $\cal{H}^{-1}=\left(q_{i,j}\right)$.
That is, $R_{n,s}^{-1}=T_{n,s}+O(n^{-1})T_{n,s}$.
I believe these "facts" can be proved using component-wise matrix perturbation theory, and I was wondering if there is some "elementary" proof and/or well-known explicit form for the entries of $R_{n,s}^{-1}$.
