convergence of a Maclaurin series I have to show that the Maclaurin series for $(1+x)^{-\frac{3}{2}}$ converges to the function for $|x| < 1 $. 
I know that I have to show that $\lim_{n\to\infty} |R_n (x)| =0 $ for $|x|<1$. 
I figured that $f^{(n+1)}=\frac{(2n)!}{2^n \times n!}\times  (1+x)^{-\frac{2n+3}{2}}$. 
From there I'm a bit stuck....
Thank very much for your help!!
Cyril
 A: Consider the Maclaurin series for $f(x) = (1+x)^{-\alpha}$ where $\alpha = 3/2$:
$$f(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!}x^k + R_n(x)$$
Taking derivatives, we have
$$f'(x) = -\alpha (1+x)^{-\alpha-1}\\ f''(x) = -\alpha(-\alpha-1) (1+x)^{-\alpha-2} = (-1)^2 \alpha(\alpha+1)(1+x)^{-\alpha-2} \\ \cdots \\ f^{(n+1)}(x) = (-1)^{n+1}\alpha(\alpha+1) \cdots (\alpha+n) (1+x)^{-\alpha-n-1}.$$
The Cauchy form of the remainder after $n$ terms in the Maclaurin expansion is
$$R_n(x) = \frac{f^{(n+1)}(\theta x)}{n!}(x - \theta x)^n x,$$
where $\theta \in (0,1)$.
In this case,
$$|R_n(x)| = \left|(-1)^{n+1} \frac{\alpha(\alpha+1) \cdots (\alpha+n) }{n!}(1+ \theta x)^{-\alpha-n-1}(x - \theta x)^n x \right| \\ = \underbrace{\alpha |x||1 + \theta x|^{-\alpha-1}}_{C_1(x)} \,\underbrace{\left|\frac{1 - \theta}{1+ \theta x}\right|^n}_{C_2(x)} \, \underbrace{\frac{(\alpha+1) \cdots (\alpha+n) }{n!}|x|^n}_{a_n|x|^n}$$
With $|x| < 1$, the first two factors $C_1(x)$ and $C_2(x)$ are bounded and independent of $n$.
Note that 
$$C_1(x) \leqslant \frac{\alpha}{|1 - \theta|^{\alpha + 1}}$$
If $0 \leqslant x < 1$, then $1 + \theta x \geqslant  1$ and $C_2(x) \leqslant |1 - \theta|^n < 1$.  On the other hand, if $-1 < x < 0$, then $1 - \theta < 1 - \theta |x| = 1 + \theta x$ and $C_2(x) < 1$.
Finally, the last factor $a_n |x|^n$ satisfies
$$\lim_{n \to \infty} \frac{a_{n+1} |x|^{n+1}}{a_n |x|^n} = \lim_{n \to \infty} \frac{\alpha + n + 1}{n+1}|x| = |x| < 1.$$
By the ratio test, the series $\sum a_n |x|^n$ converges and $a_n |x|^n \to 0$ as $ n\to \infty$.
Thus, when $|x| < 1$, we have
$$\lim_{n \to \infty} |R_n(x)| = C_1(x) C_2(x) \lim_{n \to \infty} a_n |x|^n = 0$$
