Double trouble sum

Here I wish to share my solution. :)

Here we write $2n-2k-1 =2M-3$ and $2n-2k+2=2M$ and $\Omega = \sum_{k=0}^{\infty}\sum_{k=0}^n\frac{k}{(k+1)!}\frac{(2(n-k-1)+1)!!}{(2(n-k+1))!!}=\sum_{k=0}^{\infty} \sum_{M=1}^{k}\frac{k}{(k+1)!} \frac{(2M-3)!!}{(2M)!!}$ further $\sum_{M=1}^{\infty}\sum_{k=M-1}^{\infty}\frac{k}{(k+1)!}\frac{(2M-3)!!}{(2M)!!} =\left(\sum_{k=0}^{\infty}\frac{k}{(k+1)!}\right)\left(\sum_{M=0}^{\infty}\frac{(2M-1)!!}{(2(M+1))!!}\right)$ Note that the former sum is Telescoping sum and latter sum can be expressed in terms of factorial as $\lim_{N\to\infty}\sum_{k=1}^{N} \frac{k}{(k+1)!}\left(\sum_{M=0}^{\infty}\frac{(2M-1)!!}{(2(M+1))!!}\right) =\lim_{N\to\infty} \left(1-\frac{1}{(N+1)!}\right) \beta (M)=\beta(M)$ and $\beta(M)=\sum_{M=0}^{\infty} \frac{(2M-1)!!}{(2M)!!} =\sum_{M=0}^{\infty}\frac{(2M)!}{2^M M! ( 2^{M+1}( M+1)!)}=\frac{1}{2}\sum_{M=0}^{\infty}\frac{1}{4^M(M+1)}{2M \choose M}$ Recall the Generating function for central binomial coefficient a nd hence on integrating with respect to $x$ from $0$ to $\frac{1}{4}$ we get $\int_0^{\frac{1}{4}}\frac{1}{\sqrt{1-4x}}=\int_0^{\frac{1}{4}}\sum_{n=0}^{\infty}{ 2M \choose M} x^M \Rightarrow 2\beta (M) = \int_0^{\frac{1}{4}}\frac{4}{\sqrt{1-4x}}=2$ Therefore $\beta(M)=1$ and hence is the answer.