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 $1$.
I refer you to this solution. The function $f(x)$ represented by the sum in question is
$$f(x) = \sum_{n=0}^{\infty} \frac{2^{2 n}}{\displaystyle (2 n+1) \binom{2 n}{n}} x^{2 n+1} = \frac{\arcsin{x}}{\sqrt{1-x^2}}$$
You may deduce that the radius of convergence is $1$ from the relation
$$\frac{1}{2^{2 n}} \binom{2 n}{n} \sim \frac{1}{\sqrt{\pi n}} \quad (n \to \infty)$$
which may be shown as a result of, e.g., Stirling's approximation.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Note that $\ds{\pars{2k}!! = 2^{k}\,\Gamma\pars{k + 1}}$ and $\ds{\pars{2k + 1}!! = 2^{k + 1}\,{\Gamma\pars{k + 3/2} \over \Gamma\pars{1/2}}}$. \begin{align} &\bbox[5px,#ffd]{\sum_{k = 0}^{\infty}{\pars{2k}!! \over \pars{2k + 1}!!}\,x^{2k + 1}} \\[5mm] = &\ {1 \over 2}\sum_{k = 0}^{\infty}{\Gamma\pars{k + 1} \Gamma\pars{1/2} \over \Gamma\pars{k + 3/2}}\,x^{2k + 1} \\[5mm] = & {1 \over 2}\sum_{k = 0}^{\infty}\bracks{% \int_{0}^{1}t^{k}\pars{1 - t}^{-1/2}\,\,\dd t}\,x^{2k + 1} \\[5mm] = &\ {1 \over 2}\,x\int_{0}^{1}{1 \over \root{1 - t}} \sum_{k = 0}^{\infty}\pars{x^{2}\,t}^{k}\,\dd t \\[5mm] = &\ {1 \over 2}\,x\int_{0}^{1}{1 \over \root{1 - t}} \sum_{k = 0}^{\infty}\pars{x^{2}\,t}^{k}\,\dd t \\[5mm] = &\ {1 \over 2}\,x\int_{0}^{1}{\dd t \over \root{1 - t}\pars{1 - x^{2}t}} \\[5mm] \stackrel{t\ =\ 1 - \xi^{2}}{=}\,\,\,\,\,& x\int_{0}^{1}{\dd\xi \over 1 - x^{2}\pars{1 - \xi^{2}}} \\[5mm] = &\ x\,{1 \over 1 - x^{2}}\,{\root{1 - x^{2}} \over x}\int_{0}^{1}{\pars{x/\root{1 - x^{2}}}\dd\xi \over \pars{x\xi/\root{1 - x^{2}}}^{2} + 1} \\[5mm] = & {1 \over \root{1 - x^{2}}}\arctan\pars{x \over \root{1 - x^2}} \\[5mm] = &\ \left.{\arctan\pars{\tan\pars{\theta}} \over \root{1 - x^{2}}} \,\right\vert_{\,x\ =\ \sin\pars{\theta}} = \bbx{\arcsin\pars{x} \over \root{1 - x^{2}}} \\ & \end{align}
Try to find out $\lim {a_{n+1}\over a_n}$, $a_n={1.2.\dots 2n \over 1.3.5\dots 2n+1}x^{2n+1}$
