How do I prove the following?-
$\sqrt{2}\sum_{n=0}^{\infty} \frac{\Gamma(2n+1/2)(-at)^n}{n!\Gamma(n+1/2)}$=$\frac{\sqrt{1+\sqrt{1+4at}}}{\sqrt{1+4at}}$.
I think the way to obtain the right-hand side is to show that the summation expansion of left expression can be rearranged to obtain the series expansion of the right expression. Any ideas?
Note that $$ \frac{{\Gamma (2n + \tfrac{1}{2})}}{{\Gamma (n + \tfrac{1}{2})}} = \frac{{\Gamma (n + \tfrac{1}{4})\Gamma (n + \tfrac{3}{4})}}{{\sqrt \pi \Gamma (n + \tfrac{1}{2})}}2^{2n - 1/2} = \frac{{(\tfrac{1}{4})_n (\tfrac{3}{4})_n }}{{(\tfrac{1}{2})_n }}4^n , $$ where $(a)_n$ is the Pochhammer symbol. Thus your series can be written as a Gauss hypergeometric function: $$ \sqrt 2 {}_2F_1 \left( {\tfrac{1}{4},\tfrac{3}{4},\tfrac{1}{2}, - 4at} \right). $$ Now you can use http://dlmf.nist.gov/15.4.E18 to finish the derivation.