1. ## Maclaurin series

Find the Maclaurin series (don't show that Rn(x)-->0). Also find the radius of convergence:
$(1+x)^{-3}$

So I got the first few terms: $(1)^{-3}-\frac{3(2)^{-4}}{2!}x+\frac{12(3)^{-5}}{3!}x^2-\frac{60(4)^{-6}}{4!}x^3+\frac{360(5)^{-7}}{5!}x^4-...$

I can't account for the coefficient
$\sum_{n=0}^\infty \frac{(n+1)^{-3-n}(-x)^n}{n+1}$

2. Try this: If [tex]f(u)= $\frac{1}{u}= u^{-3}$, then $f'= -3u^{-4}$, $f"= 12u^{-5}$, $f"'= -60u^{-6}$, $f""= 360u^{-7}$ and, in general, $f^{(n)}= (-1)^n \frac{(n+2)!}{2} u^{-n-2}$.

In your problem, $\frac{1}{(1+x)^{-3}}$, let u= 1+ x. x= 0 when u= 1.

3. Originally Posted by superdude
Find the Maclaurin series (don't show that Rn(x)-->0). Also find the radius of convergence:
$(1+x)^{-3}$

So I got the first few terms: $(1)^{-3}-\frac{3(2)^{-4}}{2!}x+\frac{12(3)^{-5}}{3!}x^2-\frac{60(4)^{-6}}{4!}x^3+\frac{360(5)^{-7}}{5!}x^4-...$ ???

I can't account for the coefficient
$\sum_{n=0}^\infty \frac{(n+1)^{-3-n}(-x)^n}{n+1}$

$\frac{1}{(1+x)^3} = \sum_{n=0}^{\infty} \frac{(-1)^n (n+1)(n+2) x^n}{2}$