Evaluating a series of hypergeometric functions I would like to prove (or disprove) the following statement:
$$
\sum_{n=0}^\infty \left[\frac{{}_2{\rm F}_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)}{n!}\right] = \frac{\pi}{2} \left[ \sum_{a=0}^\infty \frac{1}{4^a (a!)^2} + \frac{1}{2} \sum_{b=0}^\infty \frac{\left(\frac{1}{2}\right)^{2b}}{\Gamma\left(b+\frac{3}{2}\right)^2}\right]
$$
also can be seen as
$$
\sum_{n=0}^\infty \left[\frac{{}_2{\rm F}_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)}{n!}\right] = \frac{\pi}{2} \bigg[ I_0(1) + L_0(1) \bigg]
$$
Where $I_0(1)$ is the modified bessel function of the first kind and $L_0(1)$ is the modified struve function.
Major brownie points on the line here!
 A: I'll prove a generalization of the second expression. We first recall Gauss's theorem
$\displaystyle _2F_1(a,b;c;1)=\frac{\Gamma(c)\Gamma(c-a-b)}{\Gamma(c-a)\Gamma(c-b)}$ which implies  $$_2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)
=\frac{\Gamma(3/2)\Gamma((1+n)/2)}{\Gamma(1)\Gamma(1+n/2)}=\frac{\sqrt{\pi}}{2}\frac{\Gamma(1/2+n/2)}{\Gamma(1+n/2)}.$$ Furthermore, applying the duplication formula $\Gamma(z)\Gamma(z+1/2)=2^{1-2z}\sqrt{\pi}\,\Gamma(2z)$ for $z=1/2+n/2$ yields $\Gamma(1/2+n/2)\Gamma(1+n/2)=2^{-n}\sqrt{\pi}\,n!$. Consequently the hypergeometric function further rewrites to
$$\frac{\sqrt{\pi}}{2}\frac{\Gamma(1/2+n/2)}{\Gamma(1+n/2)}=\frac{\pi}{2}\frac{2^{-n}n!}{\Gamma(1+n/2)^2},$$ and we may write the equality
$$\sum_{n=0}^\infty\,_2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)\frac{x^n}{n!} = \frac{\pi}{2}\sum_{n=0}^\infty \frac{(x/2)^n}{\Gamma(1+n/2)^2}.$$
On the RHS, we recall that the functions involved have series definitions
$$I_0(x)=\sum_{n=0}^\infty \frac{(x/2)^{2n}}{\Gamma(1+n)^2},\quad L_0(x)=\sum_{n=0}^\infty \frac{(x/2)^{2n+1}}{\Gamma(3/2+n)^2}.$$ But, save for a factor of $\pi/2$, these are just the even and odd parts respectively of the summation just displayed. Consequently we conclude
$$\sum_{n=0}^\infty\,_2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};1\right)\frac{x^n}{n!}=\frac{\pi}{2}\left(I_0(x)+L_0(x)\right).$$ Taking $x=1$ yields the identity of interest.
A: Semiclassical has told me to post an answer to my own question using the technique i have explained to him, now for me to go any further i should state that the original function i am dealing with is:
$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x))
$ 
 therefore using the power series of a hyper geometric function i can say:
$$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) = \sum_{a=0}^{\infty} \frac{(\frac{1}{2})_a (\frac{1-n}{2})_a \cos^{2a}(x)}{(\frac{3}{2})_a a!}
$$
Having $(f(x))_a be the rising factorial therefore with the definition of a rising factorial  as using gamma functions i can state the following aswell:
$$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) =  \sum_{a=0}^{\infty} \frac{\Gamma(\frac{1}{2}+a) \Gamma(\frac{1-n}{2}+a) \Gamma(\frac{3}{2}) \cos^{2a}(x)}{\Gamma(\frac{3}{2}+a)\Gamma(\frac{1}{2}) \Gamma(\frac{1-n}{2}) a!}
$$
using the following identities i can simplify the function more
$\Gamma(\frac{1}{2}) = \sqrt{\pi}$
$\Gamma(\frac{3}{2}) = \sqrt{\pi}/2$
$ \Gamma(\frac{1}{2} + n) = \frac{(2n)! \sqrt{\pi}}{4^{n} n!}$
$$
{}_2F_1(\frac{1}{2}, \frac{1-n}{2}; \frac{3}{2};\cos^2 (x)) =  \frac{1}{2}\sum_{a=0}^{\infty}\frac{\pi(2a)!4^{a+1}(a+1)! \Gamma(\frac{1-n}{2}+a)}{ \pi 4^{a}(a!)^2 (2a+2)!  \Gamma(\frac{1-n}{2})}
$$
$$
=  \frac{1}{2}\sum_{a=0}^{\infty}\frac{4 \cos^{2a}(x) (a+1) \Gamma(\frac{1-n}{2}+a)}{ a!(2a+2)(2a+1) \Gamma(\frac{1-n}{2})}
$$
$$
=  \frac{4}{2}\sum_{a=0}^{\infty}\frac{\cos^{2a}(x) \Gamma(\frac{1-n}{2}+a)}{ 2 a!(2a+1) \Gamma(\frac{1-n}{2})}
$$
$$
= \sum_{a=0}^{\infty}\frac{\cos^{2a}(x)\Gamma(\frac{1-n}{2}+a)}{ a!(2a+1) \Gamma(\frac{1-n}{2})}
$$
Which i know looks very ugly, but when i graphed  this on Mathematica on the interval $ [0,2\pi]$ it was the same result as the function given by Semiclassical (in my specific scenario). therefore you can say that they are equal i suppose?
