# Thread: Elementary Analysis Proof (inequalities)

1. ## [SOLVED] Elementary Analysis Proof (inequalities)

Hi there,

This is my first post on the forums, so I hope people can help me out. I'll be spending a lot of time here as I'm taking two difficult university math courses, Analysis I and Honours Linear Algebra.

Anyways, here is the question, part of a suggested exercise for the Analysis class. I converted it to the form of a<b<c but that didn't help me. I'm new to the proofs and need a lot of help. Any help will be appreciated.

If $\displaystyle n \in \mathbb{N}$ prove

$\displaystyle \left |1-{\frac {n}{n+1} \right| < \epsilon \iff n> \frac{1}{\epsilon} -1 \right$

2. $\displaystyle n> \frac{1}{\epsilon} -1 \iff n + 1 > \frac{1}{\epsilon} \iff \epsilon > \frac{1}{n + 1} = \frac{1 + n - n}{n + 1} = 1 - \frac{n}{n + 1}$

Since $\displaystyle n < n + 1 \Rightarrow \frac{n}{n+1} < 1$

So the absolute value bars are not really necessary here if $\displaystyle n \ge 0$

3. That one is simple:

$\displaystyle \left| {1 - \frac{1} {{n + 1}}} \right| = \frac{1} {{n + 1}} < \varepsilon$

4. Thanks, I'm surprised I couldn't figure that one out!