Hope you're having a good day. While solving for Maclaurin series for $\arcsin(x)$, I noticed the summation of the series to be: $$ \sum_{n=1}^{\infty} \frac{x^{2n-1}((2n-3)!!)^2}{(2n-1)!}. $$ Since, $\arcsin(0.5) = \pi/6$,
$$6 \arcsin(0.5) = \pi.$$
After substituting $x = 0.5$ and multiplying by $6$, I got
$$ 6\sum_{n=1}^{\infty} \frac{\left(\frac{1}{2}\right)^{2n-1}((2n-3)!!)^2}{(2n-1)!}. $$
After running the second equation through wolfram alpha to infinty, it shows it as $\pi$.
But, I have no idea how to prove the function. I think it may have to do with binomial theorem but I don't know where to start with the proof. Any help would be appreciated.
Edit: After referring to Thomas Andrew's answer, I've managed to derive the equation
$$ \sum_{n=0}^{\infty}\frac{(2n-1)!!x^{2n+1}}{(2n)!!(2n+1)} $$
Which gives the same result as the first equation.
I've also noticed in the equation to find pi, taking
$$ \pi = 24\sum_{0}^{\infty}\frac{(2n-1)!!\sqrt{2-\sqrt{2+\sqrt{3}}}^{2n+1}}{(2n)!!(2n+1)} $$
Which gives a convergence rate of approcimately 18 digits of pi per 10 iterations
The speed of convergence increases with decrease in the value of x, which is an improvement over the first equation.
Yes, if $f(x)=\arcsin x$ then for $$f’(x)=\frac1{\sqrt{1-x^2}}=(1-x^2)^{-1/2}$$
Then we write (by the general binomial theorem):
$$f’(x)=(1-x^2)^{-1/2}=\sum_{n=0}^{\infty}(-1)^n\binom{-1/2}{n}x^{2n}$$
Then you need to find a different form for $(-1)^n\binom{-1/2}n$ and then integrate the resulting series to get a series for $f(x).$
Specifically, $$\begin{align}\binom{-1/2}{n}&=(-1)^n\frac{(2n-1)!!}{2^n n!}\\&=(-1)^n\frac{((2n-1)!!)^2}{(2n)!}\end{align}$$