Using Taylor's theorem

1. The problem statement, all variables and given/known data
Using Taylor's theorem, prove that
[tex]x-\frac{x^3}{6}<sin(x)<x-\frac{x^3}{6}+\frac{x^5}{120}[/tex] for x>0.

2. Relevant equations



3. The attempt at a solution
I know what is Taylor theorem but i haven't done some basic questions on it, i mean the easy ones. I have no idea on how should i start here.

 

Series for sin x:
[tex]sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}........[/tex]

The terms are getting smaller and smaller. For a positive x, i notice that
[tex]x-\frac{x^3}{3!}<x-\frac{x^3}{3!}+\frac{x^5}{5!}[/tex]
But i can't find my way to the answer, there's a sin x between.

 

I have never heard about "convergent alternating series" but i understand what you are saying.
As the partial sums oscillate, [itex]x-\frac{x^3}{3!}+\frac{x^5}{5!}[/itex] overestimates sin x. So i think i have reached the answer, thanks Curious for the help!

I will have to check about the convergent series, do you have any good link or should i refer Wikipedia?

 

Thank you once again Curious!