Cauchy by Definition I'm a bit confused as to how to show a sequence is Cauchy by definition. For example let $a_n = 1 - \frac{2}{\sqrt{n}}$. 
$$\left|\left( 1 - \frac{2}{\sqrt{n}}\right) - \left( 1 - \frac{2}{\sqrt{m}}\right)\right| = \left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right|$$
And let $m \geq n$. Then this is where I got lost in class: $|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}| \leq |\frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}|$.
Can someone explain why the last step is justified? Is it because $m \geq n \implies \frac{2}{\sqrt{m}} \leq \frac{2}{\sqrt{n}}$ and $\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}} \leq 0 \leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}$? But the inequality may not hold when taking absolute values of both sides...
 A: In the situation you have, with $m\geq n$, we have
$$\left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right| = \frac{2}{\sqrt{n}} - \frac{2}{\sqrt{m}} \leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}} = \left|\frac{2}{\sqrt{n}} +\frac{2}{\sqrt{n}}\right|$$
because 
$$-\frac{2}{\sqrt{m}} \leq 0 \leq \frac{2}{\sqrt{n}},$$
so adding $\frac{2}{\sqrt{n}}$ to both sides leads to the last inequality.
Alternatively,
$$\begin{align*}
\left|\frac{2}{\sqrt{m}} - \frac{2}{\sqrt{n}}\right| &\leq \left|\frac{2}{\sqrt{m}} \right| + \left|\frac{2}{\sqrt{n}}\right| &\text{(triangle inequality)}\\
&=\frac{2}{\sqrt{m}} + \frac{2}{\sqrt{n}}\\
&\leq \frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}\\
&= \left|\frac{2}{\sqrt{n}} + \frac{2}{\sqrt{n}}\right|.
\end{align*}$$
You prove a sequence $\{a_n\}$ is Cauchy "by definition" by showing that for every $\epsilon\gt 0$ there exists an $N$ (which will depend on $\epsilon$) such that for all $n,m\geq N$ we have $|a_n-a_m|\lt \epsilon$. 
A: It is a correct, though somewhat strange, deduction. Yes, one does use the fact that $m \ge n$.  Also, in general, by the Triangle Inequality, $|x-y|\le |x|+|y|$, so if $x$ and $y$ are non-negative, as is the case when $x=\frac{2}{\sqrt{m}}$ and $y=\frac{2}{\sqrt{n}}$, we have $|x-y|\le x+y$.
However, since $x$ and $y$ are both non-negative, we have $|x-y|\le \max(x,y)$, so one could have concluded that our expression is $\le \frac{2}{\sqrt{n}}$.  Perhaps the fact that the inequality obtained is less strong than the truth caused you some confusion. But certainly $\le \frac{4}{\sqrt{n}}$is plenty strong enough to prove the sequence is Cauchy. It can be helpful to train oneself to not worry too much about constants, that is, to settle for weaker inequalities in order to cut down on work.
As to "Cauchy by definition," it is not exactly by definition. However, not much  work needs to be done to verify that the sequence is Cauchy, since it is essentially built in that if $m$ and $n$ are large, then $a_m$ and $a_n$ are close to each other.  
