I'd like to make this series self contained by not relying upon outside knowledge such as Pochhammer symbols
$$\pi = 2 \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_n \left(\frac{1}{2}\right)_n}{\left(\frac{3}{2}\right)_n} \frac{1^n}{n!}$$
Tried something ugly like
$$2 \sum_{n=0}^{\infty} \frac{(\frac{1}{2}(\frac{3}{2})\cdots(\frac{1}{2}+n-1))^2}{\frac{3}{2}(\frac{5}{2})\cdots(\frac{3}{2}+n-1)n!} $$
but not sure if it's an equivalent formula or could be tidied up further?
The main reason is I'd like to write a simple C program to calculate this $\pi$ formula based on $n$.
Edit
Based upon the accepted answer, the new formula is
$$\large \pi = \sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n \cdot (n+\frac{1}{2}) \cdot n!}$$
or without the double factorial
$$\large \pi = \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n} (n!)^2} \cdot \frac{2}{2n+1}$$
Best Answer
If all you're looking for is a simplification you can use the definition $(a)_n=\Gamma(a+n)/\Gamma(a)$ and the property $\Gamma(z+1)=z\Gamma(z)$ to write $$ \begin{aligned} \pi &=2\sum_{n=0}^\infty\frac{(1/2)_n (1/2)_n}{(3/2)_n} \frac{1^n}{n!}\\ &=2\frac{\Gamma(3/2)}{\Gamma(1/2)}\sum_{n=0}^\infty\frac{\Gamma(1/2+n) }{\Gamma(3/2+n)} \frac{(1/2)_n}{n!}\\ &=\sum_{n=0}^\infty\frac{1}{1/2+n}\frac{(1/2)_n}{n!}. \end{aligned} $$
If you cannot use the Pochhammer symbol but double factorials are fine note that $$ (1/2)_n=\frac{1}{2}\cdot\frac{3}{2}\cdot\frac{5}{n}\cdots\frac{2n-1}{2}=\frac{1\cdot 3\cdot 5\cdots(2n-1)}{2^n}=\frac{(2n-1)!!}{2^n} $$ So that we have the equivalent form $$ \begin{aligned} \pi &=\sum_{n=0}^\infty\frac{(2n-1)!!}{1/2+n}\frac{2^{-n}}{n!}. \end{aligned} $$
Typing
into Mathematica does in fact return the value $\pi$.
Edit:
I do not write in C but chat GPT does. Here is C code to sum up the first eleven terms