Boundary of Equation for Arcsin(x)

Question

Continuing the study of the integral, we developed a series function for $\arcsin(x)$. However, after trying it we found that it is limited to the range $-1/2^.5$ to $1/2^.5$. The equation is below. The question is, why is it bounded to $-1/2^.5$ to $1/2^.5$

$$\arcsin(x)=\frac{x} {(1-x^2)^{1/2}}\sum_{n=0}^{\infty}\frac{x^{2n}} {(1-x^2)^{n}}\frac{(-1)^n} {(2n+1)}$$

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Answer 1

Using the Ratio Test for convergence leads to the inequality

\begin{equation} \left\vert \dfrac{x^2}{1-x^2} \right\vert<1 \end{equation} which resolves into two inequalities

\begin{align} \dfrac{1}{1-x^2}&>0\text{ and}\\ \dfrac{1-2x^2}{1-x^2}&>0 \end{align}

The solution of the first is $\vert x\vert<1$ and the solution of the second is $\vert x\vert>1$ or $\vert x\vert<\frac{\sqrt{2}}{2}$.

Thus the series converges on the interval $\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$.