How find this sum $\sum_{n=1}^{\infty}\frac{n}{[(2n)!]^2}$

Question

Find the sum $$I=\sum_{n=1}^{\infty}\dfrac{n}{[(2n)!]^2}$$

I think we can note $$\dfrac{n}{((2n)!)^2}=\dfrac{1}{2}\dfrac{2n}{((2n)!)^2)}=\dfrac{1}{2}\cdot\dfrac{1}{(2n)!\cdot(2n-1)!}$$

Answer 1

The series expansion of the Bessel function of first kind, $J_\nu(x)$, and the modified Bessel function of first kind, $I_\nu(x)=i^{-\nu}J_\nu(ix)$, are given by $$J_\nu(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\nu+1)} {\left(\frac{x}{2}\right)}^{2m+\nu}\\ I_\nu(x) = \sum_{m=0}^\infty \frac{1}{m! \, \Gamma(m+\nu+1)} {\left(\frac{x}{2}\right)}^{2m+\nu}$$ Chosing, $\nu=1$ we find that $$J_1(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, (m+1)!} {\left(\frac{x}{2}\right)}^{2m+1}\\ I_1(x) = \sum_{m=0}^\infty \frac{1}{m! \, (m+1)!} {\left(\frac{x}{2}\right)}^{2m+1}$$ Doing $I_1(x)-J_1(x)$, only the coefficients with $m$ odd will survive and we will have \begin{align}I_1(x)-J_1(x)&=\sum_{m\,\text{odd}} \frac{2}{m! \, (m+1)!} {\left(\frac{x}{2}\right)}^{2m+1}\\ &=\sum_{n=1}^\infty \frac{2}{(2n)! \, (2n-1)!} {\left(\frac{x}{2}\right)}^{4n-1}\end{align} Using your expression, we find that $$I=\frac{1}{4}(I_1(2)-J_1(2))$$