Develop the Taylor series for the function $f(x)={\frac{1-x}{\sqrt{1+x}}}$ Develop the Taylor series for the function $$f(x)={\frac{1-x}{\sqrt{1+x}}}$$ for $$ a=0$$
I have tried to differentiate it, knowing that the Taylor series have the next form: $F(x)={\frac{f^{(n)}(a)}{n!}(x-a)}$ But it's far too complicated after 2/3 derivatives

The generalized binomial theorem says that
$$
(1+x)^{-1/2}=\sum_{n\ge0}\binom{-1/2}{n}x^n \tag{1}
$$
(the radius of convergence is discussed later on).
Then
$$
\frac{1-x}{\sqrt{1+x}}=
(1-x)\sum_{n\ge0}\binom{-1/2}{n}x^n
$$
Now you can distribute and collect terms:
\begin{align}
\frac{1-x}{\sqrt{1+x}}
&=(1-x)\sum_{n\ge0}\binom{-1/2}{n}x^n
\\[6px]
&=\sum_{n\ge0}\binom{-1/2}{n}x^n-\sum_{n\ge0}\binom{-1/2}{n}x^{n+1}
\\[6px]
&=1+\sum_{n\ge1}\binom{-1/2}{n}x^n-\sum_{n\ge1}\binom{-1/2}{n-1}x^{n}
\\[6px]
&=1+\sum_{n\ge1}\left(\binom{-1/2}{n}-\binom{-1/2}{n-1}\right)x^n
\end{align}
The radius of convergence of a power series doesn't change when it's multiplied by a polynomial (verify it), so it's sufficient to look at the radius of convergence of $(1)$. With the ratio test,
$$
\left|\frac{\dbinom{-1/2}{n+1}x^{n+1}}{\dbinom{-1/2}{n}x^{n}}\right|=
\frac{n+1}{n+1/2}|x|
$$
because
$$
\binom{k}{n}=\frac{k(k-1)\dotsm(k-n+1)}{n!}
$$
so
$$
\frac{\dbinom{k}{n+1}}{\dbinom{k}{n}}=
\frac{k(k-1)\dotsm(k-n+1)}{n!}
\frac{(n+1)!}{k(k-1)\dotsm(k-n)}=
\frac{n+1}{k-n}
$$
Since the limit at $\infty$ of the ratio is $|x|$, the ratio of convergence is $1$.
