How can I evaluate this sum? $$\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2}p^n(1-p)^n$$ I know the answer from Wolframalpha, but I'm more curious about how to derive the answer. I was trying to prove that the series converges for every $p\in [0,1]$ except for $p=\frac{1}{2}$. My first attempt was to show that the series converges with a direct comparison test, that is $$\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2}p^n(1-p)^n<\sum_{n=1}^{\infty}f(n)\cdot p^n(1-p)^n$$ So I want to find a function $f(n)$ such that $\frac{(2n)!}{(n!)^2}<f(n)$ for every $n\in\mathbb{N}$. My first observation is that the sum somewhat looks like a geometric series, then I observe that $(2n)!<(2n)^{2n}$, and then I also observe that $(n!)^2<(n)^{2n}$ so we get
\begin{align*}\frac{(2n)!}{(n!)^2}&<\frac{(2n)^{2n}}{(n)^{2n}}=4^n\\ \end{align*} So we get a geometric series. But then I realized that it is completely wrong reasoning, then I was calculating $\frac{(2n)!}{(n!)^2}$ for $n=1,2,3$ to try to look for a pattern, and then I realized it seems like the inequality $\frac{(2n)!}{(n!)^2}<4^n$ seems to hold for $n=1,2,3$. To my great surprise, the inequality from my faulty reasoning is true using proof by induction so we get a geometric series $$\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2}p^n(1-p)^n<\sum_{n=1}^{\infty}(4p(1-p))^n=\frac{4p(1-p)}{1-4p(1-p)}$$
that is convergent for every $p\in [0,1]$ except for $p=\frac{1}{2}$. So I proved that the sum is less than a geometric series that converges for every $p\in [0,1]$ except for $p=\frac{1}{2}$. But my original question is still not fully answered that is proving that the series converges for every $p\in [0,1]$ except for $p=\frac{1}{2}$, I know it converges for every $p\in [0,1]$ other than $p=\frac{1}{2}$. I still don't know whether it converges or not when $p=\frac{1}{2}$, so my last resort is by using WolframAlpha to find the closed-form of that infinite sum, and finally, I see that the closed-form solution $\frac{1}{|2p-1|}-1$ is undefined when $p=\frac{1}{2}$. So my question is how can I evaluate the sum?