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Cauchy Sequences

To prove a sequence is Cauchy from the definition, an opening line of proof may go something like as follows.
Let
ϵ>0\mathsf{\epsilon > 0}
be given. It suffices to show there is a natural number
N∈N\mathsf{N\in\mathbb{N}}
such that
∣sn−sm∣≤ε\mathsf{|s_n - s_m|\leq\varepsilon}
, for all
m,n≥N.\mathsf{m,n\geq N.}
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Example 1.
Define the sequence
{sn}\mathsf{\{s_n\}}
by
sn=1+3/n\mathsf{s_n = 1 + 3/n}
for all
n∈N\mathsf{n\in\mathbb{N}}
. Prove
{sn}\mathsf{\{s_n\}}
converges by showing
{sn}\mathsf{\{s_n\}}
is Cauchy.
Ex 1 Comment
Visual
Formal Write-Up
The convergence
sn→1\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.
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Let
ϵ>0\mathsf{\epsilon > 0}
be given. It suffices to show there is a natural number
N∈N\mathsf{N\in\mathbb{N}}
such that
∣sn−sm∣≤ϵ\mathsf{|s_n-s_m|\leq\epsilon}
, for all
m,n≥N.\mathsf{m,n\geq N.}
Observe, for all indices
m\mathsf{m}
and
n\mathsf{n}
satisfying
m≥n≥N\mathsf{m\geq n\geq N}
,
​
∣sn−sm∣=∣(1+3n)−(1+3m)∣,\mathsf{|s_n - s_m| = \left| \left(1 + \dfrac{3}{n}\right) - \left(1 + \dfrac{3}{m}\right)\right|,}
​
and so
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∣sn−sm∣=3∣1n−1m∣≤3(1n+1m)≤6n,\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
m≥n\mathsf{m\geq n}
. By the Archimedean Property of
R\mathbb{R}
, there is a natural number
M∈NM\in\mathbb{N}
such that
6<Mϵ\mathsf{6 < M\epsilon}
, and so
6/M≤ϵ\mathsf{6/M \leq \epsilon}
. Thus,
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∣sn−sm∣=3∣1n−1m∣≤6min{m,n}≤6M≤ϵ,for all m,n≥M.\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
N=M\mathsf{N=M}
, by which we conclude
{sn}\mathsf{\{s_n\}}
is Cauchy.
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■\blacksquare
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