Evaluate: $\lim\limits_{r \to \infty} \frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}$ 
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:
$$\arcsin x=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}$$
 A: 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
$$
r^{1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2}=r+\frac14+O\left(\frac1r\right)
$$
A: The sum is solvable in terms of Bessel I functions, which have well-known asymptotic expansion.  Doing the sum and including the prefactor, the first two terms of the asymptotic expansion are r+1/4.  Therefore the limit is $\infty.$
Edit:
value of expression is
$$\frac{\sqrt{\pi \, r}}{2}\, e^{-r} \Big( (1+r)\,\text{I}_0(r/2) + r\, \text{I}_1(r/2) \Big) $$
Then use the following 
$$\text{I}_0(r) \sim \frac{e^r}{\sqrt{2 \pi r}} \Big( 1+ \frac{2}{8r} + ... \Big)$$
$$\text{I}_1(r) \sim \frac{e^r}{\sqrt{2 \pi r}} \Big( 1- \frac{3}{8r} + ... \Big)$$
A: On $(2,+\infty)$ gamma is increasing, so for all $n\geq1$ we have:
$$
\frac{\Gamma{(n+3/2)}}{(n!)^2} \geq
\frac{\Gamma{(n+1)}}{(n!)^2} =
\frac{n!}{(n!)^2} =
\frac{1}{n!} 
\tag{1}
$$
Thus:
$$
\sum_{n=0}^{\infty} \frac{\Gamma{(n+3/2)}r^n}{(n!)^2} \geq
\sum_{n=0}^{\infty} \frac{r^n}{n!} +\left(\Gamma{(3/2) - 1} \right) = 
e^r +\left(\Gamma{(3/2) - 1} \right) \geq
e^r - 1
\tag{2}
$$
Note that to find the lower estimation of the sum I applied $(1)$ for all terms excluding the first one ($n=0$) for which I complemented the resulting sum with $1$.
Eventually we have:
$$
\frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2} \geq
\sqrt{r} \;\frac{e^r - 1}{e^{r}}
$$
