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)!}$
 A: We have, by the extended binomial theorem:
$$ \frac{1}{\sqrt{1-x^2}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n} x^{2n},\tag{1} $$
hence by integrating termwise:
$$ \arcsin x=\sum_{n\geq 0}\frac{1}{(2n+1)4^n}\binom{2n}{n} x^{2n+1}=x\cdot\phantom{}_2 F_1\left(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^2\right)\tag{2} $$
Now we may notice that:
$$ \sum_{a+b=N}\frac{1}{(2a+1)4^a}\binom{2a}{a}\frac{1}{(2b+1)4^b}\binom{2b}{b} = \frac{4^N}{(N+1)(2N+1)\binom{2N}{N}}\tag{3}$$
We may find the Taylor series of $\arcsin^2(z)$ by exploiting the Catalan's convolution formula, the hypergeometric identity:
$$ \phantom{}_2 F_1\left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};z\right)^2 = \phantom{}_3 F_2\left(a,b,\frac{a+b}{2};\frac{a+b+1}{2},a+b;z\right)\tag{4}$$
the Lagrange's inversion theorem or the rather simple technique shown in this answer. So we have:
$$\arcsin^2(x) = \sum_{n\geq 0}\frac{4^n}{(n+1)(2n+1)\binom{2n}{n}}x^{2n+2}\tag{5}$$
and by differentiating $(5)$:

$$ \frac{\arcsin x}{\sqrt{1-x^2}} = \sum_{n\geq 0}\frac{4^n}{(2n+1)\binom{2n}{n}}x^{2n+1}. \tag{6}$$

A: Do not multiply, tackle it directly by looking at the Taylor recipe... 
Just a short answer for your check:
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\sum_{n=0}^{\infty}\frac{4^n (n!)^2}{(2n+1)!}x^{2n+1}$$
A: I would substitute $x = \sin(x)$ and simplify to $1/\cos(x)$ and proceed from there.
