Question on exponentials in inequalities

**Speed1991** · December 29th 2011, 08:32 AM

Using the fact that if $\sum|a_n|$ converges then:
$|\sum_{n=0}^{\infty}a_n|\leq\sum_{n=0}^{\infty}|a_ n|$ ,
prove that for all $z\in\bar{D}(0;1)$ (punctured disc centre 0 radius 1),
$(3-e)|z|\leq|e^z-1|\leq(e-1)|z|$ .

I haven't even got the first part yet. Ive tried manipulating e (as $\sum_{n=0}^{\infty}\frac{1}{n!}$ ) but can't get the right answer.

**girdav** · December 29th 2011, 10:03 AM

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

**Speed1991** · December 30th 2011, 05:43 AM

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

**Prove It** · December 30th 2011, 05:51 AM

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...

**Speed1991** · December 30th 2011, 08:01 AM

Sorry, I meant the first part of the inequality.

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