# Question on exponentials in inequalities

• Dec 29th 2011, 08:32 AM
Speed1991
Question on exponentials in inequalities
Using the fact that if $\displaystyle \sum|a_n|$ converges then:
$\displaystyle |\sum_{n=0}^{\infty}a_n|\leq\sum_{n=0}^{\infty}|a_ n|$,
prove that for all $\displaystyle z\in\bar{D}(0;1)$ (punctured disc centre 0 radius 1),
$\displaystyle (3-e)|z|\leq|e^z-1|\leq(e-1)|z|$.

I haven't even got the first part yet. Ive tried manipulating e (as $\displaystyle \sum_{n=0}^{\infty}\frac{1}{n!}$) but can't get the right answer.
• Dec 29th 2011, 10:03 AM
girdav
Re: Question on exponentials in inequalities
For the second inequality $\displaystyle |e^z-1|\leq \sum_{n=1}^{\infty}\frac{|z|^n}{n!}\leq \sum_{n=1}^{+\infty}\frac{|z|}{n!}=|z|(e-1)$.
• Dec 30th 2011, 05:43 AM
Speed1991
Re: Question on exponentials in inequalities
Ok, I get that. How about the first bit?
• Dec 30th 2011, 05:51 AM
Prove It
Re: Question on exponentials in inequalities
Quote:

Originally Posted by Speed1991
Ok, I get that. How about the first bit?

You aren't asked to prove the first bit, only to use it...
• Dec 30th 2011, 08:01 AM
Speed1991
Re: Question on exponentials in inequalities
Sorry, I meant the first part of the inequality.