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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?

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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.

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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.

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    What is the value of $x$ you would use for c in your above expression? And how does that expression go to zero because wouldn't there be an n+1 term on the numerator? Thanks – 2017-02-07
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    I think you are mistaken that that sequence goes to zero for $x \in (-1,0)$. But you also have the $x^{n+1}$ to help you... – 2017-02-07
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    Also you can't use $c=0$, you have $c$ between $0$ and $x$ (strictly). – 2017-02-07
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    I am still confused as to how I am supposed to use the bounds on $x$ and $c$ to show that limit goes to zero. For example, what if we take c between -1 and 0? – 2017-02-07
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    It seems you are right. This idea will not work for $x$ near $-1$. So I guess you have to use a different formula for the Maclaurin remainder to do this one? – 2017-02-07
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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 $x \in (-1,1)$.