Evaluating $\sum_{k=0}^{\infty}{(-1)^k\frac{9^ka^{-k}\Gamma(k+\frac{1}{2})}{(2k)!}} =\sqrt{\pi}e^{-\frac{9}{4a}}$

Question

How can I prove that :$$\sum_{k=0}^{\infty}{(-1)^k \, \frac{\Gamma(k+\frac{1}{2})}{(2k)!}} \, \left(\frac{9}{a}\right)^k =\sqrt{\pi} \, e^{-\frac{9}{4a}}$$

I see that $(2k)!=\Gamma(2k+1)$ and then Wolfram says $$\frac{\Gamma(k+\frac{1}{2})}{\Gamma(2k+1)}=\frac{2^{-2k}\sqrt{\pi}}{\Gamma(k+1)}$$ but I wasn't sure how to prove this part on its own either.

Answer 1

Let $a_{n} = \Gamma\left(n + \frac{1}{2}\right)$ and then consider a few $n$ values. \begin{align} a_{0} &= \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi} \\ a_{1} &= \Gamma\left(\frac{3}{2}\right) = \frac{1}{2} \, \Gamma\left(\frac{1}{2}\right) = \frac{\sqrt{\pi}}{2} \\ a_{2} &= \Gamma\left(\frac{5}{2}\right) = \frac{3}{2} \, \Gamma\left(\frac{3}{2}\right) = \frac{1 \cdot 3}{2^2} \, a_{0} = \frac{1 \cdot 2 \cdot 3 \cdot 4}{2^4 \, 1 \cdot 2} \, a_{0}\\ a_{3} &= \Gamma\left(\frac{7}{2}\right) = \frac{1 \cdot 3 \cdot 5}{2^3} \, a_{0} = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}{2^6 \, 1 \cdot 2 \cdot 3} \, a_{0}\\ \cdots &= \cdots \end{align} The pattern developed is $$a_{n} = \frac{(2 n)!}{ 2^{2 n} \, n!} \, \sqrt{\pi}. $$ This is a way to obtain the desired formula when integers are used. What happens when, given a variable $x$, is not an integer ? Well, then look to the great works of the past of Euler and Legendre. This leads to the Legendre duplication formula as given by $$ \Gamma\left(\frac{1}{2}\right) \, \Gamma(2 x ) = 2^{2 x -1} \, \Gamma(x) \, \Gamma\left(x + \frac{1}{2}\right). $$

Now, for the series being considered: \begin{align} S(x) &= \sum_{k=0}^{\infty}{(-1)^k \, \frac{\Gamma(k+\frac{1}{2})}{(2k)!}} \, \left(\frac{x}{a}\right)^k \\ &= \sqrt{\pi} \, \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k)!} \, \frac{(2 k)!}{ 2^{2 k} \, k!} \, \left(\frac{x}{a}\right)^k \\ &= \sqrt{\pi} \, \sum_{k=0}^{\infty} \frac{1}{k!} \, \left(- \frac{x}{4 \, a}\right)^{k} \\ &= \sqrt{\pi} \, e^{-x/(4 a)}, \end{align} where $$ e^{x} = \sum_{n=0}^{\infty} \frac{x^n}{n!} $$ was used.

Answer 2

$$S=\sum_{k=0}^{\infty}{(-1)^k \, \frac{\Gamma(k+\frac{1}{2})}{(2k)!}} \, \left(\frac{9}{a}\right)^k =\sqrt{\pi} \, e^{-\frac{9}{4a}}$$ Using $$\frac{\Gamma(k+\frac{1}{2})}{\Gamma(2k+1)}=\frac{2^{-2k}\sqrt{\pi}}{\Gamma(k+1)}$$ $$S=\sum_{k=0}^{\infty}\frac{(-1)^k 2^{-2k}\sqrt{\pi}(9/a)^k}{k!}= \sqrt{\pi}\sum_{k=0}^{\infty} \frac{1}{k!}(\frac{-9}{4a})^k=\sqrt{\pi}e^{\frac{-9}{4a}}.$$ Here, we have used $\sum_{k=0}^{\infty} \frac{z^k}{k!}=e^{z}.$

Answer 3

Using $$ \Gamma(x)=\int_0^\infty t^{x-1}e^{-t}dt, \sum_{k=0}^{\infty}{(-1)^k \, \frac{1}{(2k)!}}x^k=\cos(\sqrt x) $$ one has \begin{eqnarray} &&\sum_{k=0}^{\infty}{(-1)^k \, \frac{\Gamma(k+\frac{1}{2})}{(2k)!}} \, \left(\frac{9}{a}\right)^k\\ &=&\sum_{k=0}^{\infty}{(-1)^k \, \frac{1}{(2k)!}}\left(\frac{9}{a}\right)^k\int_0^\infty t^{k-\frac12}e^{-t}dt \\ &=&\int_0^\infty\sum_{k=0}^{\infty}{(-1)^k \, \frac{1}{(2k)!}}\left(\frac{9}{a}\right)^k t^{k-\frac12}e^{-t}dt \\ &=&\int_0^\infty t^{-\frac12}e^{-t}\sum_{k=0}^{\infty}{(-1)^k \, \frac{1}{(2k)!}}\left(\frac{9}{a}\right)^k t^{k}dt \\ &=&\int_0^\infty t^{-\frac12}e^{-t}\sum_{k=0}^{\infty}{(-1)^k \, \frac{1}{(2k)!}}\left(\frac{9t}{a}\right)^kdt \\ &=&\int_0^\infty t^{-\frac12}e^{-t}\cos\left(\sqrt{\frac{9t}{a}}\right)dt \\ &=&\int_0^\infty t^{-\frac12}e^{-t}\cos\left(\sqrt{\frac{9t}{a}}\right)dt \\ &=&\frac{2\sqrt a}{3}\int_0^\infty e^{-\frac{ax^2}9}\cos(x)dx \\ &=&\sqrt{\pi} \, e^{-\frac{9}{4a}}. \end{eqnarray} Here $$ \int_0^\infty e^{-\frac1ax^2}\cos(x)dx=\frac12\sqrt{a\pi}e^{-\frac a4}. $$