Radius of convergence of $f(x)=\arcsin(x)$.

Question

I am working out the series representation for the $\arcsin(x)$ function and its radius of convergence, I'm just not sure if my calculations are correct. I used the generalized binomial formula to come up with the following series representation.
\begin{align*}
\arcsin(x)=\sum_{k=0}^\infty\binom{-1/2}{k}(-1)^k\frac{x^{2k+1}}{2k+1}
\end{align*}
Now when I apply the ratio test for the radius of convergence I get the following for my my ratio.
\begin{align*}
L=\frac{a_{n+1}}{a_n}=\frac{4k^2+4k+1}{4k+6}x^2
\end{align*}
Assuming I haven't made any algebra errors am I correct that the only point of convergence is at $x=0$?

Answer 1

First, we have

$$\binom{-1/2}{k}=\frac{(-1/2)(-1/2-1)\cdots (-1/2-k+1)}{k!}\tag1$$

and

$$\binom{-1/2}{k+1}=\frac{(-1/2)(-1/2-1)\cdots (-1/2-k+1)(-1/2-k)}{(k+1)!}\tag2$$

Therefore, the ratio of $(1)$ and $(2)$ is given by

$$\frac{\binom{-1/2}{k+1}}{\binom{-1/2}{k}}=-\frac{(k+1/2)}{k+1}\tag3$$

Using $(3)$, we find that

$$\lim_{k\to\infty}\left(\frac{\binom{-1/2}{k+1}(-1)^{k+1}x^{2(k+1)+1}/(2(k+1)+1)}{\binom{-1/2}{k}(-1)^kx^{2k+1}/(2k+1)}\right)=\lim_{k\to\infty}\left(\frac{(k+1/2)(2k+1)}{(k+1)(2k+3)}\right)x^2=x^2$$