I'm interested in performing a sum, and as part of my summation, I need to express $$ \frac{1}{2n + 1}\,,\,n\in\mathbb{N}\,, $$ as a series of the form $$ \sum_{m = 0}^\infty\frac{a_m n!}{(n + m)!}\,. $$ I believe that such a feat should be possible, although I have no idea where to start! Any insight is greatly appreciated.
Edit: To clarify notation, one could have expanded as a power series: $$ \frac{1}{2n + 1} = -\sum_{m = 1}^\infty a_m\frac{1}{n^m} $$ with $a_m = (-2)^{-m}$, although this is not of the desired form. That is to say, $a_m$ must depend only on the integer $m$, and not $n$.
Edit 2: I'll also add that one can check by hand that the first several terms are $$ a_0 = 0 $$ $$ a_1 = \frac12 $$ $$ a_2 = \frac14 $$ $$ a_3 = \frac{3}{8} $$ $$ a_4 = \frac{15}{16} $$ $$ a_5 = \frac{105}{32} $$
This appears to satisfy the pattern $$ a_m = \frac{(2m-3)!!}{2^m} $$ for $m\geq 1$. It's a little annoying to see that these terms grow without bound, but perhaps my ask is simply too much!
Let $(a_m)_{m\in\mathbb{N}}$ be a solution. Taking $n\to\infty$ (which can be done termwise), we get $a_0=0$. For $|x|<1$ the series $f(x)=\sum_{m=0}^\infty a_{m+1}x^m/m!$ converges, and we have $$\frac1{2n+1}=\sum_{m=0}^\infty\frac{a_{m+1}n!}{(n+m+1)!}=\sum_{m=0}^\infty\frac{a_{m+1}}{m!}\int_0^1 x^m(1-x)^n\,dx=\int_0^1 (1-x)^n f(x)\,dx.$$ This has at most one solution $f\in L^2$ (by the density argument: the difference of any two solutions is orthogonal to each polynomial in $x$), and clearly $$f(x)=\frac1{2\sqrt{1-x}}=\frac12\sum_{m=0}^\infty\frac{(2m-1)!!}{(2m)!!}x^m$$ (here $0!!=(-1)!!=1$) is a solution. This gives the expression for $a_m$ you found.