I caught this series from the classic Method Differentialis by James Stirling:
The series is
$1+\dfrac{2}{3}x+\dfrac{8}{15}x^2+\dfrac{16}{35}x^3+\dfrac{128}{315}x^4...$
I notice that the term is formed by multiplying $\dfrac{2}{3}$ to the second term, $\dfrac{4}{5}$ to the third term, $\dfrac{6}{7}$ to the fourth term, $\dfrac{8}{9}$ to fifth term.
Is there a way to find out the closed form generating function for this power series?
On
This is the start of the expansion of $$\frac{\sin ^{-1}\left(\sqrt{x}\right)}{\sqrt{x(1-x)}}=1+\frac{2 x}{3}+\frac{8 x^2}{15}+\frac{16 x^3}{35}+\frac{128 x^4}{315}+\frac{256 x^5}{693}+\frac{1024 x^6}{3003}+O\left(x^{7}\right)$$
Written as $$\frac{\sin ^{-1}\left(\sqrt{x}\right)}{\sqrt{x(1-x)}}=\sum_{n=0}^\infty c_n x^n$$ the ratio $\frac{c_{n+1}}{c_n}$ generates the sequence $$\left\{\frac{2}{3},\frac{4}{5},\frac{6}{7},\frac{8}{9},\frac{10}{11},\frac{12}{13}, \frac{14}{15},\frac{16}{17},\frac{18}{19},\frac{20}{21},\frac{22}{23}\right\}$$ which shows the clear pattern you already noticed.
In fact,
$$c_n=\frac{\sqrt{\pi }\, \Gamma (n+1)}{2\,\Gamma \left(n+\frac{3}{2}\right)}$$
Let's write the ratio of terms as:
$$\frac{c_{n+1}}{c_n}=\frac{2(n+1)(n+1)}{2n+3} \frac{x}{n+1}$$
This makes the closed form:
$$f(x)={_2 F_1} (1,1;3/2;x)$$
Wolfram Alpha doesn't simplify it further.
That's Gauss hypergeometric function.
Sorry for being short, I'm writing from phone.