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}ϵ>0
 be given. It suffices to show there is a natural number N∈N\mathsf{N\in\mathbb{N}}N∈N
 such that ∣sn−sm∣≤ε\mathsf{|s_n - s_m|\leq\varepsilon}∣sn​−sm​∣≤ε
, for all m,n≥N.\mathsf{m,n\geq N.}m,n≥N.
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Example 1.
Define the sequence {sn}\mathsf{\{s_n\}}{sn​}
 by sn=1+3/n\mathsf{s_n = 1 + 3/n}sn​=1+3/n
 for all n∈N\mathsf{n\in\mathbb{N}}n∈N
. Prove {sn}\mathsf{\{s_n\}}{sn​}
 converges by showing {sn}\mathsf{\{s_n\}}{sn​}
 is Cauchy.
Ex 1 Comment
Visual
Formal Write-Up
The convergence sn→1\mathsf{s_n\rightarrow1}sn​→1
 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 ϵ>0\mathsf{\epsilon > 0}ϵ>0
 be given. It suffices to show there is a natural number N∈N\mathsf{N\in\mathbb{N}}N∈N
 such that ∣sn−sm∣≤ϵ\mathsf{|s_n-s_m|\leq\epsilon}∣sn​−sm​∣≤ϵ
, for all m,n≥N.\mathsf{m,n\geq N.}m,n≥N.
Observe, for all indices m\mathsf{m}m
 and n\mathsf{n}n
 satisfying m≥n≥N\mathsf{m\geq n\geq N}m≥n≥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|,}∣sn​−sm​∣=∣∣​(1+n3​)−(1+m3​)∣∣​,
​
and so
​∣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},}∣sn​−sm​∣=3∣∣​n1​−m1​∣∣​≤3(n1​+m1​)≤n6​,
​
where the first inequality is an application of the triangle inequality and the second follows from the fact m≥n\mathsf{m\geq n}m≥n
. By the Archimedean Property of R\mathbb{R}R
, there is a natural number M∈NM\in\mathbb{N}M∈N
 such that 6<Mϵ\mathsf{6 < M\epsilon}6<Mϵ
, and so 6/M≤ϵ\mathsf{6/M \leq \epsilon}6/M≤ϵ
. Thus,
​∣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.  }∣sn​−sm​∣=3∣∣​n1​−m1​∣∣​≤min{m,n}6​≤M6​≤ϵ,for all m,n≥M.
​
This shows the desired inequality holds, taking N=M\mathsf{N=M}N=M
, by which we conclude {sn}\mathsf{\{s_n\}}{sn​}
 is Cauchy.
​■\blacksquare■
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Monotone Sequences
Subsequences
Last modified 4d ago