Given$$\begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^{n}(\frac{x}{2})^{2n-\frac{1}{2}}}{n! \,\Gamma(n+\frac{1}{2})} &= \frac{(\frac{x}{2})^{\frac{-1}{2}}\sum_{n=1}^{\infty}(-1)^{n}{\bigl(\frac{-x}{4}\bigr)}^{4}}{\frac{n!(2n)!\sqrt \pi}{n!\,2^{2n}}} \\&= \sqrt{\frac{2}{x}} \sum_{n=0}^{\infty} \frac{(-1)^{n} \bigl(\frac{x^2}{4}\bigr)^n\,4^n\,n!}{n!\,(2n)!\sqrt \pi} \\&= \sqrt{\frac{2}{\pi x}} \sum_{n=0}^{\infty}\frac{(-1)^{n}\bigl(\frac{4x^2}{4}\bigr)^n}{(2n)!} \\&= \sqrt{\frac{2}{\pi x}} \sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{(2n)!}\end{aligned}$$
The expansion$$\sum_{n=0}^{\infty} \frac{(-1)^n\,x^{2n}}{(2n)!}$$is part of Taylor Series Expansion For $\cos(x)$. This series converges for all real values of $x$.
Hence,$$\sqrt{\frac{2}{x}} \cos x.$$\begin{aligned}\sum_{n=0}^{\infty}\frac{(-1)^{n}(\frac{x}{2})^{2n-\frac{1}{2}}}{n! \,\Gamma(n+\frac{1}{2})} = \sqrt{\frac{2}{x}} \cos (x-1). \end{aligned}
- Is my approach is valid here?
- Can this series be obtained through other mathematical techniques?