Maclaurin series converges to function Show that the maclaurin series for $ (1+x)^{-3/2} $ converges to the function for $|x|<1$
I'm supposed to use the remainder term $ \frac{f^{n+1}(c)x^{n+1}}{(n+1)!} $ and show that the limit of that remainder goes to zero. This has been simple for functions like $ sinx $ But it is not that easy to find the nth derivative expression for this function. Is there a simple way to this without using binomial series?
 A: For $f(x) = (1+x)^{-a}$ where $a = 3/2,$ consider the integral form of the remainder and apply the mean value theorem for integrals to obtain
$$\begin{align}R_n(x) &= \frac1{n!}\int_0^x(x-t)^nf^{(n+1)}(t) \, dt \\ &= \frac{x}{n!}(x - \theta x )^n f^{(n+1)}(\theta x) \\ &= \frac{x}{n!}(x - \theta x )^n (-1)^{n+1}a(a+1) \ldots (a+n)(1+\theta x)^{-a - n - 1}\end{align}$$
where $\theta \in (0,1),$ and 
$$|R_n(x)| = |1 + \theta x|^{-a-1}\beta_n |x|^{n+1}\left|\frac{1 - \theta}{1 + \theta x}\right|^{n},$$
where
$$\beta_n = \frac{a(a+1)\ldots (a+n)}{n!}$$
Note that if $x \geqslant 0$ then $0 < 1 - \theta < 1 + \theta x$, and if $-1 < x < 0$ then $\theta x > - \theta.$ In both cases we have $0 < 1 - \theta < 1 + \theta x$, and  
$$\left|\frac{1 - \theta}{1 + \theta x}\right|^{n} < 1.$$
Hence,
$$|R_n(x)| <  |1 + \theta x|^{-a-1}\beta_n|x|^{n+1}.$$
Note that
$$\lim_{n \to \infty} \frac{\beta_{n+1}|x|^{n+2}}{\beta_n |x|^{n+1}} = \lim_{n \to \infty} \frac{a + n + 1}{n+1}|x| = |x| < 1.$$
By the ratio test the series $\sum \beta_n |x|^{n+1}$ converges and $\beta_n |x|^{n+1} \to 0$ as $ n \to 0$.
Therefore, $R_n(x)\to 0$ as $ n \to \infty$ for all $x \in (-1,1)$.
A: It's not so bad considering the $n$th derivative.
$$
|f^{(n)}(x)| = \textstyle \frac{3}{2}\;\frac{5}{2}\;\frac{7}{2}\;\dots\;(\frac{1}{2} +n)
\;(1+x)^{-3/2-n}
< 2\cdot 3\cdots (n+1)\;(1+x)^{-3/2-n}
$$
Written this way, it should be easy to then show
$$
\frac{|f^{(n+1)}(c) x^{n+1}|}{(n+1)!}
$$
goes to zero.
added
As noted, there is $n+1$ in the numerator, so you have to show the rest goes
to zero so fast that even with factor $(n+1)$ it still goes to zero.
Of course you use $|x|<1$ and $c$ strictly between $0$ and $x$ for this.
