# Cauchy Sequences

To prove a sequence is Cauchy from the definition, an opening line of proof may go something like as follows.
Let
$\mathsf{\epsilon > 0}$
be given. It suffices to show there is a natural number
$\mathsf{N\in\mathbb{N}}$
such that
$\mathsf{|s_n - s_m|\leq\varepsilon}$
, for all
$\mathsf{m,n\geq N.}$
Example 1.
Define the sequence
$\mathsf{\{s_n\}}$
by
$\mathsf{s_n = 1 + 3/n}$
for all
$\mathsf{n\in\mathbb{N}}$
. Prove
$\mathsf{\{s_n\}}$
converges by showing
$\mathsf{\{s_n\}}$
is Cauchy.
Ex 1 Comment
Visual
Formal Write-Up
The convergence
$\mathsf{s_n\rightarrow1}$
can be verified directly from the definition of convergence, but that is not permitted by this prompt. The point here is to gain practice in a new technique.
Let
$\mathsf{\epsilon > 0}$
be given. It suffices to show there is a natural number
$\mathsf{N\in\mathbb{N}}$
such that
$\mathsf{|s_n-s_m|\leq\epsilon}$
, for all
$\mathsf{m,n\geq N.}$
Observe, for all indices
$\mathsf{m}$
and
$\mathsf{n}$
satisfying
$\mathsf{m\geq n\geq N}$
,
 ​$\mathsf{|s_n - s_m| = \left| \left(1 + \dfrac{3}{n}\right) - \left(1 + \dfrac{3}{m}\right)\right|,}$​
and so
 ​$\mathsf{|s_n - s_m| = 3\left| \dfrac{1}{n} - \dfrac{1}{m} \right|\leq 3\left(\dfrac{1}{n}+\dfrac{1}{m}\right)\leq \dfrac{6}{n},}$​
where the first inequality is an application of the triangle inequality and the second follows from the fact
$\mathsf{m\geq n}$
. By the Archimedean Property of
$\mathbb{R}$
, there is a natural number
$M\in\mathbb{N}$
such that
$\mathsf{6 < M\epsilon}$
, and so
$\mathsf{6/M \leq \epsilon}$
. Thus,
 ​$\mathsf{|s_n - s_m| = 3\left| \dfrac{1}{n} - \dfrac{1}{m} \right|\leq \dfrac{6}{\textsf{min}\{m,n\}}\leq \dfrac{6}{M}\leq\epsilon,\quad \textsf{for all}\ m,n\geq M. }$​
This shows the desired inequality holds, taking
$\mathsf{N=M}$
, by which we conclude
$\mathsf{\{s_n\}}$
is Cauchy.
 ​$\blacksquare$​