$\sum_{m=0}^{\infty}x^m\sum_{k=0}^{\infty}W_{m,k}f_k=\sum_{n=1}^{\infty}\frac{1}{n(n+1)}\sum_{k=0}^{\infty}\left(\frac{n+x}{n(n+1)}\right)^k f_k$?

Question

How do we solve

$\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n + 1)} \right)^k f_k$

for $W_{m,k}$ ?

Here is what I have attempted so far:

$\begin{array}{l} \left\{ W_{m, k} : \sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \frac{f^{(k)} (0)}{k!} \left( \frac{n + x}{n (n + 1)} \right)^k \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{n = 1}^{\infty} \sum_{k = 0}^{\infty} \frac{\frac{f^{(k)} (0)}{k!} \left( \frac{n + x}{n (n + 1)} \right)^k}{n (n + 1)} \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{m = 1}^{\infty} \sum_{k = 0}^{\infty} \frac{\left( \frac{m + x}{m (m + 1)} \right)^k}{m (m + 1)} \frac{f^{(k)} (0)}{k!} \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} \frac{\left( \frac{m + 1 + x}{(m + 1) (m + 2)} \right)^k}{(m + 1) (m + 2)} \frac{f^{(k)} (0)}{k!} \right\}\\ \left\{ W_{m, k} : \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} x^m W_{m, k} \frac{f^{(k)} (0)}{k!} = \sum_{m = 0}^{\infty} \sum_{k = 0}^{\infty} \frac{(m + 1 + x)^k}{((m + 1) (m + 2))^k (m + 1) (m + 2)} \frac{f^{(k)} (0)}{k!} \right\} \end{array}$

Answer 1

$\sum_{m = 0}^{\infty} x^m \sum_{k = 0}^{\infty} W_{m, k} f_k = \sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n (n + 1)} \right)^k f_k $

I'll naively expand the right side, not worrying about convergence.

$\begin{array}\\ \sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{n + x}{n(n+1)} \right)^k f_k &=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k(n+x)^k\\ &=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k = 0}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k\sum_{m=0}^k \binom{k}{m}x^mn^{k-m}\\ &=\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{m=0}^{\infty}\sum_{k=m}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k \binom{k}{m}x^mn^{k-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \sum_{k=m}^{\infty} \left( \frac{1}{n(n+1)} \right)^k f_k \binom{k}{m}n^{k-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{1}{n(n+1)} \right)^k \binom{k}{m}n^{k-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{n}{n(n+1)} \right)^k n^{-m}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \frac{1}{n (n + 1)} \left( \frac{1}{n+1} \right)^k\frac1{n^{m}}\\ &=\sum_{m=0}^{\infty}x^m\sum_{k=m}^{\infty}f_k\binom{k}{m}\sum_{n=1}^{\infty} \left( \frac{1}{n+1} \right)^{k+1}\frac1{n^{m+1}}\\ \end{array} $

so it looks like $ W_{m, k} =\binom{k}{m}\sum_{n=1}^{\infty} \left( \frac{1}{n+1} \right)^{k+1}\frac1{n^{m+1}} $ and $ W_{m, k} =0$ for $k < m$.