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$$\sum_{n=0}^\infty \frac{n!}{2^n\prod_{i=1}^n(1+\frac{i}{2})}$$
This question appeared while solving other alternate sums: $\sum\limits_{n=1}^\infty a_{n+1}-a_n$ and the thing is that it converges searching via: $\lim\limits_{n\to\infty} S_n=\lim\limits_{n\to\infty}a_{n+1}-a_1$
The problem is to express the first sum as an alternate sum, so I need to find a formula for $a_n$ and the rest is easy. I've solved the others problems using partial fraction decomposition for the first expression.
As @Mark pointed out, you're problem can be reexpressed as.
$$\sum\limits_{n=0}^{\infty}\frac{n!}{2^n \frac{(n+2)!}{2^{n+1}}}$$
$$\sum\limits_{n=0}^{\infty}\frac{2 n!}{(n+2)!}$$
$$\sum\limits_{n=0}^{\infty}\frac{2}{(n+1)(n+2)}$$
$$2 \times \left(\sum\limits_{n=0}^{\infty}\frac{1}{n+1} - \sum\limits_{n=0}^{\infty}\frac{1}{n+2}\right)$$
Making a change of variables
$$2 \times \left(\sum\limits_{k=1}^{\infty}\frac{1}{k} - \sum\limits_{k=2}^{\infty}\frac{1}{k}\right)$$
$$2 \times \left(\sum\limits_{k=1}^{\infty}\frac{1}{k} - \left( \sum\limits_{k=1}^{\infty}\frac{1}{k} - 1 \right) \right)$$
$$2$$
We have $$ S=\sum_{n=0}^{+\infty} \frac{n!}{2^n \prod_{i=1}^n \left(1+\frac{i}{2} \right)}, $$ but we know $$ \prod_{i=1}^n \left(1+\frac{i}{2} \right)=\frac{\Gamma(n+3)}{2^{(n+1)}} = \frac{(n+2)!}{2^{(n+1)}}. $$ So, we have $$S=\sum_{n=0}^{+\infty} \frac{n!}{2^n \frac{(n+2)!}{2^{(n+1)}}} = 2 \sum_{n=0}^{+\infty} \frac{n!}{(n+2)!} = 2\sum_{n=0}^{+\infty} \left( \frac{1}{n+1}-\frac{1}{n+2} \right) = 2$$
