Cauchy Criterion on the series $\frac{1}{n}$ I have seen the statement a million times and I know it implies convergence but I've never actually seen how to use it.
It seems much harder than proving a limit since there is an $m$ and an $n$. Can someone give an example, or even use this Criterion to show that $\frac{1}{n}$ is a convergent sequence?
Suppose I have reduced the Cauchy Criterion inequality in a problem to
$\frac{n}{m}$, $m\gt n$, and $\frac{n}{m} > \epsilon$.
If I want to find a lower bound for $n$, can I choose anything I want that will satisfy the above inequalities?
 A: To prove that $\frac{1}{n}$ is cauchy, say you are given $\epsilon>0$. Then, you want to find an $N$ such that if $n,m\geq N$ then $|\frac{1}{m}-\frac{1}{n}|<\epsilon$, this reads to $$|\frac{1}{m}-\frac{1}{n}|=|\frac{n-m}{mn}|\leq|\frac{n}{mn}|=\frac{1}{m}$$ where above say $n\geq m$ so if both $m$ and $n$ are bigger than $\frac{1}{\epsilon}$, then you would have that $|\frac{1}{m}-\frac{1}{n}|<\epsilon$, I agree that this problem does not really require you to use Cauchy, and it is easier to do it directly. However, in analysis you are going to have problem were you have to use this a lot. 
A: It's very easy to prove that any convergent sequence is Cauchy, because, if $x = \lim x_n$, then $|x_n-x_m|<|x_n-x|+|x_m-x|$.  Those cases don't give us more than than the standard limit definition.
The more interesting cases are when we don't immediately "know" what the limit is gonna be.
Let $0<\alpha<1$.  If $f:\mathbb R\rightarrow \mathbb R$ (or $f:[a,b]\rightarrow [a,b]$ for some $a,b$,) such that $|f(x)-f(y)|<\alpha |x-y|$ for all $x,y$.
Given any $x_0$, define $x_{n+1}=f(x_n)$.  Then $\{x_0,x_1,...,x_n,...\}$ is easily shown to be Cauchy, but it's much less obvious that it converges.
(The interesting side-affect of this example is the the limit $x=\lim x_n$ has the property that $f(x)=x$.  But we can show that $f$ can only have one "fixed" value by our condition above.  If $x\neq y$ and $f(x)=x$ and $f(y)=y$, then $|f(x)-f(y)|=|x-y|>\alpha |x-y|$.
For example, if $\beta$ is any real, then $f(x)=\beta+\alpha \arctan x$ has this property.  It's not obvious what $\{x_0,f(x_0),f(f(x_0)),...\}$ converges to, but we know it converges.
Similarly, if $\{b_1,b_2,...,b_n,...\}$ is a convergent increasing sequence of real numbers, and $\{a_1,a_2,...,a_n,...\}$ has the property that for all $n$, $|a_{n+1}-a_n|\leq b_{n+1}-b_n$ then $\{a_n\}$ is easily shown to be Cauchy, even though we might have no idea what it converges to.
A: Let us stick for a second with the sequence $x_n=1/n$. We already know this converges to zero, to it can help understand what is going on. If you take any interval around zero, i.e. of the form $(0-\delta,0+\delta)$, then all the terms in the sequence, except finitely many, are inside the interval (draw a picture if you don't see it). In particular, if $n_1$ is the biggest index such that $x_n$ is not in the interval, we have that for any $n,m>n_1$, 
$$
\left|\frac1n-\frac1m\right|<2\delta.
$$
So, eventually, we get this Cauchy relation. 
The reason we take two elements in the sequence, is because we want to apply this idea to sequences where we don't know if there is a limit. And it turns out (that's exactly Cauchy's criterion) that a sequence of real numbers is convergent if and only if it is Cauchy. 
A: In practice, you will want to eliminate either $n$ or $m$ before you bound your difference by epsilon- for the same reason that, for example, it is easier to find a solution to $\sin(2x)=\dfrac1 2$ rather than $\sin(x)\cos(x)=\dfrac 1 4$. The latter might not seem relevant since the trig expressions are equal, but you'd prefer to have an easier expression to work with. Hope that helps.
