# Innequality prove question

1. Dec 21, 2008

### transgalactic

2. Dec 21, 2008

### HallsofIvy

Staff Emeritus
You need a little bit more. It is true that
$$x- \frac{x^3}{3!}+ \frac{x^5}{5!}> x- \frac{x^3}{3!}$$
but you have to fit
$$sin(x)= \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
into that. Notice that because this is an alternating series with decreasing terms, the entire "tail" (all terms past "n") is less than the difference between the n-1 and n terms.

3. Dec 21, 2008

### transgalactic

what is the steps for the solution?

4. Dec 21, 2008