This is a follow-up question for the one described here.
Evaluate $$\lim\limits_{r \to \infty} r-\frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}=1/4$$
There is a related but not complete answer here. But, I was not sure how to complete it. Also, one might write $\frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}$ in terms of Bessel Functions and then use the expansion of the Bessel Functions.
I think that, in the previous post, skbmoore's answer gives the key $$\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}\,r^n}{(n!)^2}=\frac{\sqrt{\pi }}{2} e^{r/2} \left((r+1) I_0\left(\frac{r}{2}\right)+r I_1\left(\frac{r}{2}\right)\right)$$ $$y= r-\frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}=r-\frac{\sqrt{\pi r}}{2} e^{-r/2} \left((r+1) I_0\left(\frac{r}{2}\right)+r I_1\left(\frac{r}{2}\right)\right)$$
Now, using equation $9.7.1$ in page $377$ of the Handbook of Mathematical Functions by Abramowitz and Stegun we have, for large values of $z$ $$I_0(z)\sim \frac{e^z}{\sqrt{2 \pi z } }\left(1+\frac{1}{8 z}+\frac{9}{128 z^2}+\frac{75}{1024 z^3}+O\left(\frac{1}{z^4}\right)\right)$$ $$I_1(z)\sim \frac{e^z}{\sqrt{2 \pi z } }\left(1-\frac{3}{8 z}-\frac{15}{128 z^2}-\frac{105}{1024 z^3}+O\left(\frac{1}{z^4}\right)\right)$$ which lead to $$y=-\frac{1}{4}-\frac{1}{32 r}-\frac{3}{128 r^2}-\frac{75}{256 r^3}+O\left(\frac{1}{r^4}\right)$$
Consider $r=10$ : the exact value would be $y\approx -0.253411$ while the above expansion leads to $-\frac{12987}{51200}\approx -0.253652$.
