Evaluate: $\lim\limits_{r \to \infty} \frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2}$

Question

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: $$\arcsin x=\sum _{n=0}^{\infty }{\frac {(2n)!}{4^{n}(n!)^{2}(2n+1)}}x^{2n+1}$$

Answer 1

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 $$ \begin{align} \Gamma\!\left(n+\tfrac32\right) &\ge\frac{\Gamma(n+1)^{3/2}}{\Gamma(n)^{1/2}}\\ &=n^{1/2}\Gamma(n+1)\\[9pt] &=n!\sqrt{n}\tag2 \end{align} $$

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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 $$ \begin{align} r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2} &\ge r^{-1/2}e^{-r}\sum_{n=0}^\infty\frac{\sqrt{n}\,r^n}{n!}\\ &\ge r^{-1/2}\left[e^{-r}\sum_{n=0}^\infty\frac{n r^n}{n!}\right]^{3/2}\left[e^{-r}\sum_{n=0}^\infty\frac{n^2\,r^n}{n!}\right]^{-1/2}\\[9pt] &=r^{1/2}\left(r+1\right)^{-1/2}\tag4 \end{align} $$

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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 $\infty$.

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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 $$ r^{1/2}e^{-r}\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\frac32\right)r^n}{(n!)^2}=r+\frac14+O\left(\frac1r\right) $$

Answer 2

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 $$\text{I}_0(r) \sim \frac{e^r}{\sqrt{2 \pi r}} \Big( 1+ \frac{2}{8r} + ... \Big)$$ $$\text{I}_1(r) \sim \frac{e^r}{\sqrt{2 \pi r}} \Big( 1- \frac{3}{8r} + ... \Big)$$

Answer 3

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: $$ \frac{\sqrt{r}}{e^{r}}\sum_{n=0}^{\infty}\frac{\Gamma{(n+3/2)}r^n}{(n!)^2} \geq \sqrt{r} \;\frac{e^r - 1}{e^{r}} $$