@skbmoore
Prove that $$\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}\,r^n}{(n!)^2}=\frac{\sqrt{\pi }}{2} e^{r/2} \left((r+1) I_0\left(\frac{r}{2}\right)+r I_1\left(\frac{r}{2}\right)\right)$$
This is a follow-up question of this.
I have been trying to verify this, but I only did not know how to. I searched for the expansion of $I_0$ and $I_1$ something with an exponential function, but nothing found. Any idea?
This can be solved in a CAS, but to make it interesting, let's do most of the work ourselves. By properties of the gamma function $$ \frac{1}{\sqrt{\pi}} \sum_{n=0}^\infty \frac{\Gamma(n+3/2)}{n!^2}r^n = \sum_{n=0}^\infty (n+1/2)\frac{\Gamma(n+1/2)}{\Gamma(1/2)} \frac{r^n}{n!^2} =\big(r\,\frac{d}{dr}+\frac{1}{2}\big) \underbrace{ \sum_{n=0}^\infty \frac{r^n}{n!}\,2^{-2n}\binom{2n}{n}}_{:=S(r)}$$ It is assumed the OP can do the derivatives and algebra. My part of the derivation is through when I can show $$ (*) \quad S(r) = \sum_{n=0}^\infty \frac{r^n}{n!}\,2^{-2n}\binom{2n}{n} = e^{r/2}\,I_0(r/2)$$ where $I_0$ is the modified Bessel function. This can probably be shown by a (Cauchy) power series product, but I chose to reduce it to the well-known integral relationship $$ I_0(x)=\frac{1}{\pi} \int_0^\pi \exp{\big(-x\,\cos(t)\big)} dt$$ Use another well-known relationship (I believe derivable from a trig substitution into the beta integral) for the central binomials.
$$ \frac{2}{\pi} \int_0^{\pi/2}\! \sin^{2n}(t) \,dt = 2^{-2n}\binom{2n}{n}$$
Insert this into the definition for $S(r)$ and interchange $\sum$ and $\int.$ The sum is simply the exponential and the expression so obtained is $$S(r)=\frac{2}{\pi} \int_0^{\pi/2} \exp{\big( r\,\sin^2(t)\,\big)} dt= \frac{2}{\pi} \int_0^{\pi/2} \exp{\big(\frac{r}{2}\,(1-\cos(2t))\,\big)} dt $$ by a trigonometric identity. Finally, by rescaling the integral $$S(r)=e^{r/2} \, \frac{1}{\pi} \int_0^{\pi}\exp{\big(\!-\frac{r}{2}\,\cos(t)\big)} dt$$ which completes the proof of (*).
