Evaluate:
$$\lim\limits_{r \to \infty} \frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}$$
My effort:
\begin{aligned}\Gamma \left({\tfrac {1}{2}}+n\right)&={(2n)! \over 4^{n}n!}{\sqrt {\pi }}
\end{aligned}
Therefore,
$$ \frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}= \frac{\sqrt{r \pi}}{4 e^{r}}\sum_{n=0}^{\infty}\frac{{(2n+2)! }}{4^{n}(n+1)! (n!)^2}r^n = \frac{\sqrt{\pi}}{4 e^{r}}\sum_{n=0}^{\infty}\frac{{(2n+2)! }}{4^{n}(n+1)! (n!)^2}{(\sqrt{r})}^{2n+1}, $$
which is very similar to:
$$\arcsin x=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}$$
Best Answer
Due to the log-convexity of $\Gamma(x)$, we have $$ \begin{align} \Gamma\!\left(n+\tfrac32\right) &\le\Gamma(n+1)^{1/2}\Gamma(n+2)^{1/2}\\ &=n!\sqrt{n+1}\tag1 \end{align} $$ and $$ \begin{align} \Gamma\!\left(n+\tfrac32\right) &\ge\frac{\Gamma(n+1)^{3/2}}{\Gamma(n)^{1/2}}\\ &=n^{1/2}\Gamma(n+1)\\[9pt] &=n!\sqrt{n}\tag2 \end{align} $$
Therefore, Cauchy-Schwarz says $$ \begin{align} r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2} &\le r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\sqrt{n+1}\,r^n}{n!}\\ &\le r^{-1/2}\left[e^{-r}\sum_{n=0}^\infty\frac{r^n}{n!}\right]^{1/2}\left[e^{-r}\sum_{n=0}^\infty\frac{(n+1)\,r^n}{n!}\right]^{1/2}\\[9pt] &=r^{-1/2}(r+1)^{1/2}\tag3 \end{align} $$ and $$ \begin{align} r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2} &\ge r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\sqrt{n}\,r^n}{n!}\\ &\ge r^{-1/2}\left[e^{-r}\sum_{n=0}^\infty\frac{n r^n}{n!}\right]^{3/2}\left[e^{-r}\sum_{n=0}^\infty\frac{n^2\,r^n}{n!}\right]^{-1/2}\\[9pt] &=r^{1/2}\left(r+1\right)^{-1/2}\tag4 \end{align} $$
The Squeeze Theorem along with $(3)$ and $(4)$ shows that $$ \lim_{r\to\infty}r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2}=1\tag5 $$ This is a different limit than asked for, but it shows that the limit in the question diverges to $\infty$.
Further Estimates
It can easily be shown that for integer $k\ge0$, $$ \begin{align} e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+1+k\right)r^n}{(n!)^2} &=e^{-r}\sum_{n=0}^\infty\frac{\overbrace{(n+1)(n+2)\cdots(n+k)}^{n^k+\frac{k(k+1)}2n^{k-1}+O\left(n^{k-2}\right)}r^n}{n!}\\ &=e^{-r}\sum_{n=0}^\infty\frac{\overbrace{n(n-1)\cdots(n-k+1)}^{n^k-\frac{k(k-1)}2n^{k-1}+O\left(n^{k-2}\right)}r^n}{n!}\\ &+k^2e^{-r}\sum_{n=0}^\infty\frac{\overbrace{n(n-1)\cdots(n-k+2)}^{n^{k-1}+O\left(n^{k-2}\right)}r^n}{n!}\\ &+O\Bigg(e^{-r}\sum_{n=0}^\infty\frac{\overbrace{n(n-1)\cdots(n-k+3)}^{n^{k-2}+O\left(n^{k-3}\right)}r^n}{n!}\Bigg)\\[9pt] &=e^{-r}\sum_{n=k}^\infty\frac{r^n}{(n-k)!}\\ &+k^2e^{-r}\sum_{n=k-1}^\infty\frac{r^n}{(n-k+1)!}\\ &+O\left(e^{-r}\sum_{n=k-2}^\infty\frac{r^n}{(n-k+2)!}\right)\\[6pt] &=r^k\left(1+\frac{k^2}r+O\left(\frac1{r^2}\right)\right) \end{align} $$ Plugging in $k=\frac12$ gives $$ r^{1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2}=r+\frac14+O\left(\frac1r\right) $$