<s>[INST] $1-\cos (x) \leq \frac{x^2}{2} + \frac{x^3}{6}$, for $x > 0$, using Taylor expansion I want to solve this problem using Taylor expansions.
I tried
\begin{align*}
1 - \cos (x) = 1 - \left( \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} +R_{2n}(x) \right) \;,
\end{align*}
where $R_{2n}(x)$ is the remainder function.
For $n=2$, this yields
\begin{align*}
1 - \cos (x) = 1 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} + R_4(x) \right) = \frac{x^2}{2} - \frac{x^4}{24} - o(x^4) \;.
\end{align*}
Is it now possible to just state
\begin{align*}
|o(x^4)| \leq \frac{x^4}{24} \; ,
\end{align*}
and hence
\begin{align*}
\frac{x^2}{2}-\frac{x^4}{24}-o(x^4)\leq \frac{x^2}{2} \leq \frac{x^2}{2} + \frac{x^3}{6} \; ?
\end{align*}
 [/INST] For $x>0$,
$$\cos x > 1 -\frac{x^2}2$$
$$\begin{aligned}1-(\cos x) &< 1-\left(1 -\frac{x^2}2\right)\\
1-\cos x&< \frac{x^2}2 +\frac{x^3}6\end{aligned}$$
</s>