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