Summation involving Gamma function

Question

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?

Gary · Answer

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.

Summation involving Gamma function

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