Showing that the Taylor series of $f(x)=\frac{1}{\sqrt{1+x}}$ converges to $f$ on (I think) the interval (-1,1]. Let $f(x)=\frac{1}{(1+x)^{1/2}}$. Supposed to find the Taylor series $Tf$ of $f$ around $0$, and show that it converges to $f$ on $[-1,1)$, (although I suspect there's a misprint in the book, and that this is supposed to be $(-1,1]$, since $f$ is not defined in $x=-1$). 
If $f(x)=T_nf(x)+R_nf(x)$, where $T_nf(x)$ denotes the $n$th Taylor polynomial, then there's a result in my book which says, in short, that if the $(n+1)$-th derived $f^{(n+1)}$ of $f$ is such that there's a number $M$ with $|f^{(n+1)}(t)|\leq M$ for all $t\in(-1,1]$, then $|R_n f(x)|\leq \frac{M}{(n+1)!}|x|^{n+1}.$ My plan was to use this to show that $|R_n f(x)|\rightarrow 0$ as $n\rightarrow \infty$ for $x\in(-1,1]$. Can't seem to find any such $M$, however, and actually such an $M$ seems unlikely to me, seeing as, if I'm not mistaking, $|f^{(n+1)}(t)|=\frac{3*5*\cdots*(2n+1)}{2^{n+1}|1-t|^{\frac{2n+3}{2}}}.$ For doesn't this grow big when $t$ is close enough to $-1$?
I may however very well be on an entirely wrong track here...with the $R_n f$ argument and stuff.
Very thankful for any help.
 A: The proof uses a different strategy for the cases $0\le x<1$ and $-1<x<0$. Let $f(x)=(1+x)^\alpha$, $\alpha\in\mathbb{R}$, and let
$$
\binom{\alpha}{n}=\frac{\alpha(\alpha-1)\dots(\alpha-n+1)}{n!}.
$$
Then
$$
\frac{f^{(n)}(x)}{n!}=\binom{\alpha}{n}(1+x)^{\alpha-n}.
$$
The Taylor series for $f$ is
$$
\sum_{n=0}^\infty\binom{\alpha}{n}x^n.
$$
Using the ratio test we find that the radius of convergence of the series is $1$. This implies that for any $r\in(0,1)$
$$
\lim_{n\to\infty}\left|\binom{\alpha}{n}\right|r^n=0.\tag{1}
$$
The remainder is
$$
R_n(x)=f(x)-\sum_{k=0}^n\binom{\alpha}{k}x^k=
\binom{\alpha}{n+1}(1+\theta)^{\alpha-n-1}x^{n+1},
$$
where $\theta$ is a real number between $0$ and $x$. We want to show that $\lim_{n\to\infty}R_n(x)=0$ for $x\in(-1,1]$
First case: $0\le x<1$. Then
$$
|(1+\theta)^{\alpha-n-1}|\le\max(1,2^\alpha)\qquad\forall n\ge0.
$$
The desired result follows from (1).
Second case: $-1/2<x<0$. Then
$$
|(1+\theta)^{\alpha-n-1}|\le C_\alpha(1-|x|)^{-n-1}\qquad\forall n\ge0
$$
and
$$
|(1+\theta)^{\alpha-n-1}x^{n+1}|\le C_\alpha\Bigl(\frac{|x|}{1-|x|}\Bigr)^n,\quad\frac{|x|}{1-|x|}<1.
$$
The result follows again from (1)
Third case: $-1<x<1/2$. This seems to be more complicated. The integral form of the remainder might be useful. I'll keep looking into it.
Fourth case: $x=1$. This follows from Abel's Theorem.
A: This is a binomial expansion. See this reference. Every mathematician will probably solve this problem by mean of complex analytic functions. Actually, you can choose a continuous branch of $z \in \mathbb{C} \mapsto (1+z)^\alpha$, which turns out to be analytic for $|z|<1$. This is the argument by Ahlfors, Complex Analysis, page 180.
If you prefer a real-analysis approach, you can


*

*Prove that $\sum_n \begin{pmatrix} \alpha \\ n \end{pmatrix} x^n$ converges when $-1<x<1$;

*differentiate term by term this series and discover that it solves a differential equation with an initial condition. Then, by uniqueness of solution, the series is your $f$.


Important edit: I have found a complete and yet elementary proof. Please refer to this excerpt from Hijab, Introduction to calculus and analysis.
A: The binomial theorem states:$$
(1+x)^\alpha = \sum_{n=0}^\infty \begin{pmatrix}\alpha \\ n\end{pmatrix} x^n
$$
Meaning that the Taylor coefficient $c_n$ in your case is:
$$\begin{pmatrix}-1/2 \\ n\end{pmatrix}$$
So the ratio test gives the radius of convergence as:
$$r= \lim_{n\rightarrow\infty}\left|\frac{c_{n+1}}{c_n}\right|=1$$
