Calculate the following sequence $\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{n!},\; \alpha >0$

Calculate the following sequence $$\sum_{n=0}^{+\infty }\left ( -\dfrac{1}{4\alpha } \right )^{n}\dfrac{ (2n)!}{n!},\; \alpha >0$$

Solution 1:

Note that $\dfrac{(2n)!}{n!}=(2n)(2n-1)\cdots(n+1)\ge n^n$, therefore, $$ \left|\,\left(-\frac1{4\alpha}\right)^n\frac{(2n)!}{n!}\,\right| \ge\left|\,\frac{n}{4\alpha}\,\right|^n\tag1 $$ Thus, the terms of the series do not go to $0$, so the series diverges.

However, if as suggested by Lucian, this is supposed to be $$ \sum_{k=0}^\infty\binom{2n}{n}\left(-\frac1{4\alpha}\right)^{\large n}\tag2 $$ then $$ \begin{align} \binom{2n}{n} &=2^n\frac{(2n-1)!!}{n!}\tag{3a}\\ &=4^n\frac{\left(n-\frac12\right)!}{n!\left(-\frac12\right)!}\tag{3b}\\ &=4^n\binom{n-\frac12}{n}\tag{3c}\\[3pt] &=(-4)^n\binom{-\frac12}{n}\tag{3d} \end{align} $$ Explanation:
$\text{(3a)}$: $(2n)!=(2n-1)!!\,2^nn!$
$\text{(3b)}$: $(2n-1)!!=2^n\frac{\left(n-\frac12\right)!}{\left(-\frac12\right)!}$
$\text{(3c)}$: write ratio as a binomial coefficient$\\[9pt]$
$\text{(3d)}$: negative binomial coefficient

Thus, $$ \begin{align} \sum_{n=0}^\infty\binom{2n}{n}\left(-\frac1{4\alpha}\right)^{\large n} &=\sum_{n=0}^\infty\binom{-\frac12}{n}\frac1{\alpha^n}\tag{4a}\\[3pt] &=\left(1+\frac1\alpha\right)^{-1/2}\tag{4b}\\[6pt] &=\sqrt{\frac{\alpha}{\alpha+1}}\tag{4c} \end{align} $$

Solution 2:

This is a binomial series. And also, it's $(n!)^2$.