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.

Re: Question on exponentials in inequalities

For the second inequality $\displaystyle |e^z-1|\leq \sum_{n=1}^{\infty}\frac{|z|^n}{n!}\leq \sum_{n=1}^{+\infty}\frac{|z|}{n!}=|z|(e-1)$.

Re: Question on exponentials in inequalities

Ok, I get that. How about the first bit?

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...

Re: Question on exponentials in inequalities

Sorry, I meant the first part of the inequality.