Lecture 26:

Cauchy Sequences

Definition. A sequence \(X=(x_{n})\) of real numbers is said to be a Cauchy sequence if for every \(\varepsilon>0\) there exists a natural number \(H\) such that\[|x_{n} - x_{m}|<\varepsilon\quad\text{for all }m,n\geq H.\]

Pop Quiz. Which of the following is a correct restatement of the definition of a Cauchy sequence?

  1. There is some distance \(\varepsilon\) so that all of the terms of the sequence \(X\) are within \(\varepsilon\) of each other. 
  2. There is a natural number \(H\) so that if \(n\geq H\) and \(m\geq H,\) then the distance between \(x_{m}\) and \(x_{n}\) will be less than \(\varepsilon\) for all \(\varepsilon>0\).
  3. If \(\varepsilon\) is a positive distance, then for all numbers \(m\) and \(n\) sufficiently large, the terms \(x_{n}\) and \(x_{m}\) will be within \(\varepsilon\) of each other.
  4. The terms of the sequence \(X\) are converging to \(x_{n}\).

Pop Quiz. Which of the following is a correct restatement of the definition of a Cauchy sequence?

  1. There is some distance \(\varepsilon\) so that all of the terms of the sequence \(X\) are within \(\varepsilon\) of each other. 
  2. There is a natural number \(H\) so that if \(n\geq H\) and \(m\geq H,\) then the distance between \(x_{m}\) and \(x_{n}\) will be less than \(\varepsilon\) for all \(\varepsilon>0\).
  3. If \(\varepsilon\) is a positive distance, then for all number \(m\) and \(n\) sufficiently large, the terms \(x_{n}\) and \(x_{m}\) will be within \(\varepsilon\) of each other.
  4. The terms of the sequence \(X\) are converging to \(x_{n}\).

Example. The sequence \((\frac{1}{n})\) is a Cauchy sequence.

Proof. Let \(\varepsilon>0\) be given. By the Archimedean property there is a natural number \(H\) such that \(H\geq \frac{2}{\varepsilon}+1.\) If \(m,n\geq H\), then we have

\[m\geq H>\frac{2}{\varepsilon}\quad\text{and}\quad n\geq H>\frac{2}{\varepsilon}\]

which implies

\[\frac{1}{m}<\frac{\varepsilon}{2} \quad\text{and}\quad \frac{1}{n}<\frac{\varepsilon}{2}\]

By the triangle inequality,

\[\left|\frac{1}{n} - \frac{1}{m}\right|\leq \frac{1}{n} + \frac{1}{m}<\varepsilon.\quad\Box\]

Example. The sequence \((\frac{2n}{n+1})\) is a Cauchy sequence.

Proof. Let \(\varepsilon>0\) be given. Since \(\lim X = 2\) we know that there is some \(H\in\N\) such that \[\left|\frac{2n}{n+1}-2\right|<\frac{\varepsilon}{4}\quad\text{for all }n\geq H.\] If \(n,m\geq H\), then

\[\left|\frac{2n}{n+1} - \frac{2m}{m+1}\right| = \left|\frac{2n}{n+1} - 2 + 2 - \frac{2m}{m+1}\right|\qquad\qquad\]

\[\leq \left|\frac{2n}{n+1} - 2\right| + \left|2 - \frac{2m}{m+1}\right|\]

\[<\frac{\varepsilon}{4} + \frac{\varepsilon}{4}=\frac{\varepsilon}{2}<\varepsilon.\quad\Box\]

Theorem. If \(X=(x_{n})\) is a convergent sequence of real numbers, then \(X\) is a Cauchy sequence.

Proof. Let \(\varepsilon>0\) be given. Let \(x=\lim X\). There exists a natural number \(K\in\N\) such that

\[|x_{n}-x|<\frac{\varepsilon}{4}\quad\text{for all }n\geq K.\]

If \(m,n\geq K\) then we have

\[|x_{n} - x_{m}| = |x_{n} - x + x - x_{m}|\leq |x_{n}-x|+|x-x_{m}|<\frac{\varepsilon}{4}+\frac{\varepsilon}{4} = \frac{\varepsilon}{2}<\varepsilon.\]

\(\Box\)

Example. The sequence \(\big(1+(-1)^{n}\big)\) is not a Cauchy sequence.

 

Set \(\varepsilon=\frac{1}{2}\). Let \(H\in\N\) be arbitrary. Let \(n\) be an even number greater than \(H\) and let \(m\) be an odd number greater than \(H\). It follows that \(x_{n} = 2\) and \(x_{m} = 0\), thus

\[|x_{n} - x_{m}|=|2-0|=2>\frac{1}{2} = \varepsilon.\]

What does it mean to say that \(X=(x_{n})\) is not a Cauchy Sequence?

You tell me!

There exists a number \(\varepsilon>0\) such that for every \(H\in\N\) there are natural numbers \(n\) and \(m\) such that \(n,m\geq H\) and \(|x_{n}-x_{m}|\geq \varepsilon.\)

Example. Define the sequence \(S = (s_{n})\) by

\[s_{n} = \sum_{k=1}^{n}\frac{1}{k}\quad\text{for all }n\in\N.\]

Notice that

\[s_{n+1}-s_{n} = \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}+\frac{1}{n+1}\right) - \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)\]

\[ = \frac{1}{n+1}\]

So, for very large \(n\), the terms \(s_{n}\) and \(s_{n+1}\) are very close to each other.

But, is \((s_{n})\) Cauchy?

Proposition. If \(|x_{n}-x_{m}|<1\) for all \(n,m\in\N\), then \((x_{n})\) is bounded.

Theorem. If \((x_{n})\) is a Cauchy sequence, then \((x_{n})\) is bounded. 

We will start next time with:

Practice Problem.

Proof. Set \(M = 1+|x_{1}|\). If \(n\in\N\), then by the triangle inequality
\[|x_{n}| = |x_{n}-x_{1}+x_{1}|\leq |x_{n} - x_{1}| + |x_{1}|<1+|x_{1}|=M.\quad \Box\]

Read Section 3.5 in the text