Determine the power series representation of the function $f(x)=\sqrt{(4+x)^3}$ and indicate the radius of convergence I want to find a representation of the function mentioned above, so I took into account that:
$$f(x)=\sqrt{(4+x)^3}=8\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$$ and developing the binomial series for $\left(1+\frac{x}{4}\right)^{\frac{3}{2}}$ we have to
$$\left(1+\frac{x}{4}\right)^{\frac{3}{2}}=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdot\cdot\cdot$$
Now, to obtain the terms of the original series simply multiply by 8; now, how can I express the function as a series, since I have only managed to express that as a sum of terms
Any help is appreciated
 A: Starting from
$$ \frac{1}{\sqrt{1-x}}=\sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n}\,x^n \tag{1}$$
which is important and should be treated as a fundamental "building block", in my opinion
one gets by termwise integration
$$ \sqrt{1-x} = \sum_{n\geq 0}\frac{\binom{2n}{n}}{4^n(1-2n)}\,x^n \tag{2}$$
and by integrating again
$$ (1-x)^{3/2} = \sum_{n\geq 0}\frac{3\binom{2n}{n}}{4^n(2n-1)(2n-3)}\,x^n. \tag{3}$$
By replacing $x$ with $-x/4$ we get
$$ \left(1+\frac{x}{4}\right)^{3/2} = \sum_{n\geq 0}\frac{3\binom{2n}{n}(-1)^n}{16^n(2n-1)(2n-3)}\,x^n \tag{4}$$
so
$$\boxed{ \sqrt{(4+x)^3} = \sum_{n\geq 0}\frac{24\binom{2n}{n}(-1)^n}{16^n(2n-1)(2n-3)}\,x^n} \tag{5}$$
A: According to the binomial series expansion we obtain
\begin{align*}
\left(1+\frac{x}{4}\right)^{\frac{3}{2}}&=1+\frac{3}{2}\left(\frac{x}{4}\right)+\frac{\frac{3}{4}}{2!}\left(\frac{x}{4}\right)^2+\frac{\frac{-3}{8}}{3!}\left(\frac{x}{4}\right)^3+\cdots\\
&=1+\binom{3/2}{1}\frac{x}{4}+\binom{3/2}{2}\left(\frac{x}{4}\right)^2+\binom{3/2}{3}\left(\frac{x}{4}\right)^3+\cdots\\
&\,\,\color{blue}{=\sum_{j=0}^\infty\binom{3/2}{j}\left(\frac{x}{4}\right)^j}
\end{align*}
