<s>[INST] Proving $x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} + \frac{x^5}{120} ~~ \forall x \in \Bbb R^+$ using Taylor's expansion I'm trying to prove 

$$x-\frac{x^3}{6} < \sin(x) < x - \frac{x^3}{6} +  \frac{x^5}{120} ~~ \forall x \in \Bbb R^{+}$$

I wrote $\sin(x)=x-x^3/3!+x^5/5!-x^7/7!+\dots$ and then the expression for $\sin(x)-(x-\frac{x^3}{6})$ i.e. $x^5/5!-x^7/7!+x^9/9!+\dots$. 
I don't see why $x^5/5!-x^7/7!+x^9/9!+\dots$ should be positive for all positive real $x$. Any idea?
 [/INST] All of them fit in this infinite sequence of inequalities:
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots \ge 0 \ \forall x \in \mathbb{R}  \\
\frac{x^3}{3!} - \frac{x^5}{5!} + \frac{x^7}{7!} -\cdots > 0 \ \forall x>0 \\
\frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots > 0  \ \forall x \in \mathbb{R^*}\\
\cdots \cdots \\
$$
It's enough to show the first sum is $\ge 0$ and $>0$ for $0<|x|< \epsilon$, and notice that the derivative of any of those sums is the previous sum. The first inequality follows from the identity of power series 
$$\frac{x^2}{2!} - \frac{x^4}{4!} + \frac{x^6}{6!} - \cdots= 2 \left( \sum_{k \ge 0}\,(-1)^k\frac{x^{2k+1}}{2^{2k+1}(2k+1)!}\right)^2$$ which one can check directly. 
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