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$\sum_{n=1}^{\infty } \frac{4^{n}}{n!\left ( n+2 \right )!} = \sum_{n=1}^{\infty } \frac{4^{n}}{ \left ( 2n+1 \right )!B\left ( n+1,n+1 \right ) \left ( n+1 \right ) \left ( n+2 \right )}$

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.

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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 $k$.

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  • $\begingroup$ As often on this site, this "answer" is at most a rewriting of the question since the special function at hand is defined as the sum of this series... $\endgroup$ – Did Feb 6 at 20:06
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    $\begingroup$ Not necessarily. There are many ways to define Bessel functions, e.g. as solutions of Bessel's differential equation, or by an integral representation. $\endgroup$ – Robert Israel Feb 6 at 20:19
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    $\begingroup$ Perhaps more importantly, the answer makes it clear that the OP can stop wasting time looking for an "elementary" expression for this sum using methods that could never lead to a Bessel function. $\endgroup$ – Robert Israel Feb 6 at 20:24

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