According to the Euler's duplication formula: $$ \Gamma(z) \Gamma(z+\frac{1}{2}) = 2^{1-2z} \sqrt{\pi} \Gamma(2z) \therefore $$ $$ \Gamma(2z) = \frac{\Gamma(z) \Gamma(z+\frac{1}{2}) 2^{2z}}{2\sqrt{\pi}} $$ So call $b = (\frac{n}{2}-a)$ and $c = (\frac{n}{2}) $ $$ \frac{\Gamma(b) \Gamma(2c)}{\Gamma(c)\Gamma(2b)} = \frac{\Gamma(b) \frac{\Gamma(c)\Gamma(c+\frac{1}{2}) 2^{2c}}{2\sqrt{\pi}}}{\Gamma(c)\frac{\Gamma(b)\Gamma(b+\frac{1}{2}) 2^{2b}}{2\sqrt{\pi}}} $$ When this is simplified it is equal to: $$ \frac{\Gamma(c+\frac{1}{2}) 2^{2(c-b)}}{\Gamma(b+\frac{1}{2})} = \frac{\Gamma(\frac{n+1}{2}) 2^{2a}}{\Gamma(\frac{n+1}{2} - a)} $$
So the reason i am simplifying this gamma function is so that i can put an expression that is an infinite sum inside of an infinite sum into a simpler form. So lets call the expression i am using this in f(x). Also note that mathematica was able to graph f(x) before this simplification.But when i applied what i have shown above to f(x) (lets call this expression g(x)) and tried to graph g(x) using Mathematica, Mathematica automatically simplified the g(x) into one infinite sum and then proceeded to say that there is a whole bunch errors, yet it simplified to a whole different function that would actually help me if i did this right and it turns out that f(x) = g(x) = this whole different function. Are there any flaws in my work?