find the sum of$\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!}$ $\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!} = \sum_{n=1}^{\infty } \frac{4^{n}}{ \left ( 2n+1 \right )!B\left ( n+1,n+1 \right ) \left ( n+1 \right ) \left ( n+2 \right )}$
Given,  $F(n) = \frac{4^{n}}{ \left ( 2n+1 \right )!B\left ( n+1,n+1 \right )  }$
then,  $\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!} = \sum_{n=1}^{\infty } \frac{F\left ( n \right )}{ \left ( n+1 \right )\left ( n+2 \right ) } =  \frac{F(1)}{2} + \frac{F(2) - F(1)}{3} + \frac{F(3) - F(2)}{4} + \frac{F(4) - F(3)}{5} + \cdots \cdots $
finally, I couldn't solve it. Please guide me how to deal with this problem. Thank you in advance.

The sum is $$-\frac{1}{2}+\frac{{I}_{2}(4)}{4}$$
where $I_2$ is a modified Bessel function of order $2$.  More generally,
$$ \sum_{n=0}^\infty \frac{x^n}{n!(n+k)!} = \frac{I_k(2 \sqrt{x})}{x^{k/2}}$$
for nonnegative integers $k$.
