I gotta show that

$\displaystyle \frac{1}{2!}-\frac{x^2}{4!} < \frac{1-cos(x)}{x^2} < \frac{1}{2!}$ for x not equal to 0

I know that $\displaystyle |S-S_{n}| < a_{n+1}$ and $\displaystyle S_{n-1} < S < S_{n}$ or $\displaystyle S_{n} < S < S{n-1}$ depending on which one is bigger value and smaller value

Using the MacLaurin series for cos x and applying $\displaystyle \frac{1-cos (x)}{x^2}$ to it, I got

$\displaystyle \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n+1)!} = \frac{1}{2!} - \frac{x^2}{4!} + \frac{x^4}{6!}.....$

so clearly

$\displaystyle \frac{1}{2!} - \frac{x^2}{4!} < \,or\, > \frac{1 - cos(x)}{x^2} < \,or\, > \frac{1}{2!}$

what's the best way to show which of the two it is?

$\displaystyle \frac{1}{2!} - \frac{x^2}{4!} < \frac{1}{2!} $ for all values of x except x=0

^^ just use this argument?