Proof of arcsin taylor series coefficients

Question

I found out by looking at the taylor series of $\arcsin(x)={\sin}^{-1}(x)=x+\frac{1}{6}x^3+\frac{3}{40}x^5+\frac{5}{112}x^7+...$ that it could most likely be written like this: $$\arcsin(x)=\sum_{k=0}^\infty \frac{((2k-1)!!)^2}{(2k+1)!} x^{2k+1}=\sum_{k\in odd} \frac{((k-2)!!)^2}{k!} x^k$$ With $n!!=n(n-2)(n-4)...$, the double factorial of n. This series seems to work and I checked the first 10 termos. But I couldn't find this series on the internet for the $\arcsin(x)$ function.

I would like to know if there already is a proof for this fact. I also don't know if this series is even correct and approximates the $\arcsin(x)$ around $x=0$.

Answer 1

Based on an answer in Maclaurin expansion of $\arcsin x$

From the fact that $ (1+z)^\alpha=\sum\limits_{k=0}^{+\infty}\frac{\alpha(\alpha-1)\ldots(\alpha-k+1)}{k!}z^k $ with $z=-u$ and $\alpha=-\frac12$ we get: \begin{align} \frac 1{\sqrt{1-u}}&=1+\frac12 u+\frac12\cdot\frac32\frac{u^2}{2!}+\frac12\cdot\frac32\cdot\frac52\frac{u^3}{3!}++\frac12\cdot\frac32\cdot\frac52\cdot\frac 72\frac{u^4}{4!}+\dotsm \\ &=1+\frac12 u+\frac{1\cdot3}{2^2\,2!}u^2 +\frac{1\cdot3\cdot 5}{2^3\,3!}u^3 + +\frac{1\cdot3\cdot 5\cdot 7}{2^4\,4!}u^4\dotsm , \end{align} and substituting $u=x^2$, then integrating term by term by using the fact that $\arcsin x = \int_0^x \frac{dt}{(1-t^2)^{1/2}}$, we obtain: \begin{align} \arcsin x&= x+ \frac12 \frac{x^3}3+\frac{1\cdot3}{2^2\,2!}\frac{x^5}5 +\frac{1\cdot3\cdot 5}{2^3\,3!}\frac{x^7}7 + +\frac{1\cdot3\cdot 5\cdot 7}{2^4\,4!}\frac{x^9}9+\dotsm \\[1ex] &=x+ \frac12 \frac{x^3}3+\frac{1\cdot3}{2\cdot 4}\frac{x^5}5 +\frac{1\cdot3\cdot 5}{2\cdot4\cdot 6}\frac{x^7}7 +\frac{1\cdot3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}\frac{x^9}9+\dotsm \\[1ex] &=\sum_{n=1}^\infty\frac{(2n-1)!!}{(2n)!!}\frac{x^{2n+1}}{2n+1}\\[1ex] &=\sum_{n=1}^\infty\frac{(2n-1)!! \cdot (2n-1)!!}{(2n)!! \cdot (2n-1)!! \cdot (2n+1)}x^{2n+1} \\[1ex] &=\sum_{n=1}^\infty\frac{((2n-1)!!)^2}{(2n+1)!}x^{2n+1}.\\[1ex] \end{align}