Summing series with factorials in How do you sum this series?
$$\sum _{y=1}^m \frac{y}{(m-y)!(m+y)!}$$
My attempt:
$$\frac{y}{(m-y)!(m+y)!}=\frac{y}{(2m)!}{2m\choose m+y}$$
My thoughts were, sum this from zero, get a trivial answer, take away the first term. But actually I don't think this will work very well.
This question was originally under probability, but the problem is that I can't sum a series and really has nothing to do with probability (reason for the first comment)

For example, one can write
\begin{align}
\sum_{y=0}^m\frac{y}{(m-y)!(m+y)!}
&=
\sum_{k=0}^m\frac{m-k}{k!(2m-k)!}
\\
&=\frac{m}{(2m)!}\sum_{k=0}^m{2m \choose k}-\frac{1}{(2m-1)!}\sum_{k=1}^{m}{2m-1\choose k-1}
\\
&=
\frac{m}{2(2m)!}\left[{2m\choose m}+\sum_{k=0}^{2m}{2m \choose k}\right]-\frac{1}{(2m-1)!}\sum_{k=0}^{m-1}{2m-1\choose k}
\\
&=
\frac{m}{2(2m)!}\left[{2m\choose m}+\left(1+1\right)^{2m}\right]-\frac{1}{2(2m-1)!}\sum_{k=0}^{2m-1}{2m-1\choose k}
\\
&=
\frac{m}{2(2m)!}{2m\choose m}+\frac{m\cdot 2^{2m}}{2(2m)!}-\frac{2^{2m-1}}{2(2m-1)!}
\\
&=
\frac{m}{2(2m)!}{2m\choose m}\;.
\end{align}
All we have used in the way is that $\displaystyle{n\choose k} ={n\choose n-k}$ and that $\displaystyle(1+1)^n=\sum_{k=0}^n{n\choose k}$.
