<s>[INST] $\frac{x^5}{5!}- \frac{x^7}{7!}+ \frac{x^9}{9!}+\dots>0$ holds for $x>0$ Show that: 
$\sin x> x- \dfrac{x^3}{6} ~~\forall ~x>0$
Attempt: 
Let $y = \sin x - x+\dfrac{x^3}{6}$
We have to prove that y is increasing for $x >0$
$y' = \cos x -1 + \dfrac{x^2}{2}$
$y'' =-\sin  x+x$
$y''>0$ for all $x>0$. 
$\implies y'$ is increasing for all $x>0$
$\implies y$ is an increasing function with $y(0+)>0$
Hence proved. 
Attempt 2: 
Writing the Taylor expansion of $\sin x$, 
we get this inequality to be proven: 
$\dfrac{x^5}{5!}- \dfrac{x^7}{7!}+ \dfrac{x^9}{9!}+...>0$
Is it possible to prove it? I tried to (by rearranging) but couldn't.
 [/INST] Consider any alternating series $a_1-a_2+a_3...$ with $a_n >0$ and $a_n$ decreasing to $0$. If the series is absolutely convergent then  we can group the terms as $(a_1-a_2)+(a_3-a_4)+...$ and the sum is greater than $a_1-a_2$ because the other terms are non-negative. Your question is a special case of this. Attempt 2 also works by the same method. Details: $x^{5} /5! -x^{7} /7! >0$ if $x <\sqrt {42}$ Similarly $x^{7} /7! -x^{9} /9! >0$, etc for such $x$ so we get $\sin x > x- x^{3} /3!$. It remains to see $\sin x > x-x^{3} /3!$ when $x \geq \sqrt {42}$. But here $x-x^{3} /3!<-1 \leq \sin x$ since $x^{3} >6x>6x-6$.
</s>