Evaluate this power series Evaluate the sum
$$x+\frac{2}{3}x^3+\frac{2}{3}\cdot\frac{4}{5}x^5+\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{6}{7}x^7+\dots$$
Totally no idea. I think this series may related to the $\sin x$ series because of those missing even powers. Another way of writing this series:
$$\sum_{k=0}^{\infty}\frac{(2k)!!}{(2k+1)!!}x^{2k+1}.$$

In this answer, I mention this identity, which can be proven by repeated integration by parts:
$$
\int_0^{\pi/2}\sin^{2k+1}(x)\;\mathrm{d}x=\frac{2k}{2k+1}\frac{2k-2}{2k-1}\cdots\frac{2}{3}=\frac{1}{2k+1}\frac{4^k}{\binom{2k}{k}}\tag{1}
$$
Your sum can be rewritten as
$$
f(x)=\sum_{k=0}^\infty\frac1{(2k+1)}\frac{4^k}{\binom{2k}{k}}x^{2k+1}\tag{2}
$$
Combining $(1)$ and $(2)$, we get
$$
\begin{align}
f(x)
&=\int_0^{\pi/2}\sum_{k=0}^\infty\sin^{2k+1}(t)x^{2k+1}\,\mathrm{d}t\\
&=\int_0^{\pi/2}\frac{x\sin(t)\,\mathrm{d}t}{1-x^2\sin^2(t)}\\
&=\int_0^{\pi/2}\frac{-\,\mathrm{d}x\cos(t)}{1-x^2+x^2\cos^2(t)}\\
&=-\frac1{\sqrt{1-x^2}}\left.\tan^{-1}\left(\frac{x\cos(t)}{\sqrt{1-x^2}}\right)\right]_0^{\pi/2}\\
&=\frac1{\sqrt{1-x^2}}\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\\
&=\frac{\sin^{-1}(x)}{\sqrt{1-x^2}}\tag{3}
\end{align}
$$

Radius of Convergence
This doesn't appear to be part of the question, but since some other answers have touched on it, I might as well add something regarding it.
A corollary of Cauchy's Integral Formula is that the radius of convergence of a complex analytic function is the distance from the center of the power series expansion to the nearest singularity. The nearest singularity of $f(z)$ to $z=0$ is $z=1$. Thus, the radius of convergence is $1$.
