<s>[INST] Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$ Determine the Taylor series expansion of function $f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}}$.
$f(x) = \frac{\arcsin(x)}{\sqrt{1-x^2}} = \arcsin(x)\frac{1}{\sqrt{1-x^2}}$
It is known: 
(1.) $\arcsin(x) = \sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
(2.) $\frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1-x^2}} = \frac{d}{dx}(\sum_{n=0}^\infty\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}) = \sum_{n=0}^\infty\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
(3.) In third step I multiplied these two series, but I am not sure whether it is correct:
$\arcsin(x)\frac{1}{\sqrt{1-x^2}} = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
EDIT:
How did you get this sum $\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$ ?
And, in case I have to determine the product of the (some other) series, which one of these should I write? 
(a) $\sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m+1)!!x^{2m+2}}{2^{m+1}(m+1)!})\frac{(2n-1)!!x^{2n+1}}{2^nn!(2n+1)}$
or 
(b) $ = \sum_{n=0}^\infty(\sum_{m=0}^n\frac{(2m-1)!!x^{2m+1}}{2^mm!(2m+1)})\frac{(2n+1)!!x^{2n+2}}{2^{n+1}(n+1)!}$
 [/INST] Do not multiply, tackle it directly by looking at the Taylor recipe... 
Just a short answer for your check:
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$$
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