Find the closed form for the following power series I wish to find the closed form for this power series:
$1+\dfrac{1}{6}x+\dfrac{3}{40}x^2+\dfrac{5}{112}x^3+\dfrac{35}{1152}x^4...$
I have been able to spot the that the second term is the first term multiply with $\dfrac{1 \cdot 1}{2 \cdot 3}$, the third term is the second term multiply with $\dfrac{3 \cdot 3}{4 \cdot 5}$, the fourth term is the third term multiply with
$\dfrac{5 \cdot 5}{6 \cdot 7}$, the fifth term is the fourth term multiply with
$\dfrac{7 \cdot 7}{8 \cdot 9}$ and so on.
How do I start to find the closed form for this series? It is taken from James Stirling's Methodus Differentialis. The author doesn't present a closed form

As noted in the comments we have for $|x|\lt1$
$$\arcsin{(x)}=x+\frac16x^3+\frac3{40}x^5+\frac5{112}x^7+\dots$$
Hence the provided power series is just
$$1+\frac16x+\frac3{40}x^2+\frac5{112}x^3+\dots=\begin{cases}1&x=0\\\frac{\arcsin{(\sqrt{x})}}{\sqrt{x}}&x\ne0\end{cases}$$
