Lecture 27:

The Cauchy Criterion

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

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

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

 

Proof. Since \((x_{n})\) is Cauchy, there is some \(H\in\N\) such that

\[|x_{n}-x_{m}|<1\quad\text{for all }n,m\geq H\]

Note that if \(n\geq H,\) then

\[|x_{H} - x_{n}|<1,\]

and hence

\[-1+x_{H}<x_{n}<1+x_{H}.\]

Set

\[M = \max\{|x_{1}|,|x_{2}|,\ldots,|x_{H-1}|,|1+x_{H}|,|-1+x_{H}|\}.\]

Then we have

\[|x_{n}|\leq M\quad\text{for all }n\in\N,\]

that is, \((x_{n})\) is bounded. \(\Box\)

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

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

Let's look at the subsequence \((s_{2^{m}})\):

\[s_{2^{1}} = s_{2} = 1 + \frac{1}{2}\]

\[s_{2^{2}} = s_{4} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \geq 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{4} = 1+\frac{2}{2}\]

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

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

\[s_{2^{m}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+ \cdots + \frac{1}{2^{m}}\]

Example continued. Hence, given any \(M\in\R\), by the Archimedean Property we can find a natural number \(m>2M-2\).

This implies \(\frac{m}{2}+1>M\), thus

\[s_{2^{m}}\geq 1+\frac{m}{2}>M.\]

This shows that \((s_{n})\) is not bounded, and hence \((s_{n})\) is not Cauchy.

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

 

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

Assume \(X = (x_{n})\) is a Cauchy sequence. We have already seen that \(X\) is bounded.

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

Assume \(X = (x_{n})\) is a Cauchy sequence. We have already seen that \(X\) is bounded. By the Bolzano-Weierstrass theorem there is a subsequence \(X' = (x_{n_{k}})\) that converges to some number \(x^{\ast}\).

 

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

Assume \(X = (x_{n})\) is a Cauchy sequence. We have already seen that \(X\) is bounded. By the Bolzano-Weierstrass theorem there is a subsequence \(X' = (x_{n_{k}})\) that converges to some number \(x^{\ast}\).

Let \(\varepsilon>0\) be given. 

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

Assume \(X = (x_{n})\) is a Cauchy sequence. We have already seen that \(X\) is bounded. By the Bolzano-Weierstrass theorem there is a subsequence \(X' = (x_{n_{k}})\) that converges to some number \(x^{\ast}\).

Let \(\varepsilon>0\) be given. Since \(X\) is a Cauchy sequence there is a number \(M\in\N\) such that \[|x_{n}-x_{m}|<\frac{\varepsilon}{2}\quad\text{for all }n,m\geq M.\] 

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

Assume \(X = (x_{n})\) is a Cauchy sequence. We have already seen that \(X\) is bounded. By the Bolzano-Weierstrass theorem there is a subsequence \(X' = (x_{n_{k}})\) that converges to some number \(x^{\ast}\).

Let \(\varepsilon>0\) be given. Since \(X\) is a Cauchy sequence there is a number \(M\in\N\) such that \[|x_{n}-x_{m}|<\frac{\varepsilon}{2}\quad\text{for all }n,m\geq M.\] Since \(X'\) converges to \(x^{\ast}\) there is a number \(K\in\N\) such that \[|x_{n_{k}}-x^{\ast}|<\frac{\varepsilon}{2}\quad\text{ for all }k\geq K.\]

 

Theorem (Cauchy Convergence Criterion). A sequence of real numbers is convergent if and only if it is a Cauchy sequence.

Proof. We have already proved that convergent sequences are Cauchy.

Assume \(X = (x_{n})\) is a Cauchy sequence. We have already seen that \(X\) is bounded. By the Bolzano-Weierstrass theorem there is a subsequence \(X' = (x_{n_{k}})\) that converges to some number \(x^{\ast}\).

Let \(\varepsilon>0\) be given. Since \(X\) is a Cauchy sequence there is a number \(M\in\N\) such that \[|x_{n}-x_{m}|<\frac{\varepsilon}{2}\quad\text{for all }n,m\geq M.\] Since \(X'\) converges to \(x^{\ast}\) there is a number \(K\in\N\) such that \[|x_{n_{k}}-x^{\ast}|<\frac{\varepsilon}{2}\quad\text{ for all }k\geq K.\]

Fix some \(K_{0}\geq K\) such that \(n_{K_{0}}\geq M\). If \(n\geq M\) then we have

\[|x_{n} - x^{\ast}| = |x_{n} - x_{n_{K_{0}}} + x_{n_{K_{0}}} - x^{\ast}| \leq |x_{n} - x_{n_{K_{0}}}|+ |x_{n_{K_{0}}} - x^{\ast}|<\varepsilon.\ \Box\]

Example. Consider the sequence \((x_{n})\) defined by

\[x_{1} = 1,\quad x_{2} = 3,\quad\text{and}\quad x_{n+2}=\frac{1}{2}(x_{n}+x_{n+1})\text{ for all }n\in\N\]

Example continued. 

\[|x_{n}-x_{m}| = |x_{n} - x_{n+1} + x_{n+1} - x_{n+2} + x_{n+2} -\cdots +x_{m-1}-x_{m}|\]

\[\leq |x_{n} - x_{n+1}| + |x_{n+1} - x_{n+2}| + \cdots + |x_{m-1}-x_{m}|\]

Example continued. Hence, if we are given any \(\varepsilon>0\), we can choose \(M\in\N\) large enough that

\[\frac{1}{2^{M-3}}<\varepsilon\]

If \(m,n\in\N\) such that \(m,n\geq M\) and \(m>n\), then

\[|x_{n}-x_{m}| < \frac{1}{2^{m-3}}\leq \frac{1}{2^{M-3}}<\varepsilon.\]

This shows that \((x_{n})\) is a Cauchy sequence. By the Cauchy Convergence Criterion the sequence \((x_{n})\) is convergent.

 

Practice Problem. Define the sequence \((a_{n})\) where

\[a_{1} = 1\quad\text{and}\quad a_{n} = a_{n-1} + \frac{1}{2^{n}}\quad\text{for all  }n\geq 2\]

Use a proof very similar to the one in the previous example to show that \((a_{n})\) is a Cauchy sequence.