Show directly that $x_n = {n+1 \over n}$ is a Cauchy sequence

Question

I started out my proof by stating that for $\epsilon > 0$, there is a natural number $N(\epsilon/2)$ such that $$\text{if $n,m > N(\epsilon/2)$ then $|x_m - x_n| < \epsilon$.}$$ Then I worked it out until I ended up with ${1 \over m} + {1 \over n}\le {\epsilon \over 2} + {\epsilon \over 2} = \epsilon$. My teacher wrote back that I needed to use the Archimedean Property because "Cauchyness" can't be assumed and I'm confused on what I need to do.

Answer 1

Note that $$ \frac1n+\frac1m=\varepsilon $$ is likely to fail for most choices of $\varepsilon$, since the left-hand-side is rational.

As your teacher says, you cannot assume that your sequence is Cauchy. Here is one way to argue. You have $$ \left|x_m-x_n\right|=\left|\frac{m+1}m-\frac{n+1}n\right|=\left|\frac1m-\frac1n\right|\leq\frac1{\min\{m,n\}}. $$ The last step assures us that we can deal with each of $m,n$ at a time. We want the right-hand-side to be less than $\varepsilon$. This implies that we need both $m,n>1/\varepsilon$.

So if you take $N(\varepsilon)$ be an integer greater than $1/\varepsilon$, then whenever $m,n>N(\varepsilon)$ you'll have that $|x_m-x_n|<\varepsilon$.