11.4 Cauchy sequences

Cauchy sequences

In order to prove that \((a_n)\) is convergent:

Find \(l\) (the candidate)
\(\forall \varepsilon >0,\) we need to find \(n_0(\varepsilon)\) such that \(n>n_0\Rightarrow|a_{n}-l|<\varepsilon\)

It is usually hard to take \(\varepsilon\), then Cauchy sequences give us another method to deal with the sequences. When sequence is not monotonic, then we cannot use the theorem to ensure it has a limit

Alternative use Cauchy sequences

Definition

We say that \((a_n)\) is a Cauchy sequence if \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|a_{n_1}-a_{n_2}|<\varepsilon\)

Example

Show that \((x_n := 1/n)\) is a Cauchy sequence.

Proof

Let \(\epsilon > 0\) and pick \(N \in \mathbb{N}\) such that \(2/N < \epsilon\) by the Archimedean Property.

Now suppose \(n, m \geq N\) with \(n, m \in \mathbb{N}\), we have \(|x_n - x_m| = |1/n - 1/m| \leq |1/n| + |1/m| \leq 2/N < \epsilon\)

By definition, \((x_n)\) is a Cauchy sequence.

Theorem

A sequence is convergent if and only if it is a Cauchy sequence.

Analysis

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Proof

\(\Rightarrow\)) We know that \(a_n\) is convergent. Let \(l\) be the limit.

Hence: \(\forall \varepsilon>0,\exists \tilde{n}_0(\varepsilon_1):n>\tilde{n}_0\Rightarrow|a_n-l|<\varepsilon_1\)

We want: Given \(\varepsilon>0\) we need to find \(n_0\): \(n_1,n_2>n_0\Rightarrow|a_{n_1}-a_{n_2}|<\varepsilon\)

Let \(\varepsilon>0\), put \(\varepsilon_1=\frac{\varepsilon}{2}\). This provides \(\tilde{n}_0\) such that \(\forall\frac{\varepsilon}{2}>0,\exists\tilde{n}_0(\varepsilon_1):n>\tilde{n}_0\Rightarrow|a_{n}-l|<\frac{\varepsilon}{2}\)

Let \(n_0=\tilde{n}_0\), if \(n_1,n_2>n_0=\tilde{n}_0\Rightarrow\begin{cases}|a_{n_1}-l|<\varepsilon_1=\frac{\varepsilon}{2}\\ |a_{n_2}-l|<\varepsilon_1=\frac{\varepsilon}{2}\end{cases}\)

Then \(|a_{n_1}-a_{n_2}|=|a_{n_1}-l+l-a_{n_2}|\leq|a_{n_1}-l\left|+\right|l-a_{n_2}|<\varepsilon\)

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Lemma: Any Cauchy sequence \(a_n\) is bounded.

Proof

We know that \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|a_{n_1}-a_{n_2}|<\varepsilon\)

Take \(\varepsilon=1\), then \(\forall\varepsilon>0,\exists n_0=n_0(\varepsilon)\) such that \(n_1,n_2>n_0\Rightarrow|a_{n_1}-a_{n_2}|<1\)

Let \(n_1=n>n_0\) and \(n_2=n_0+1>n_0\Rightarrow|a_{n}-a_{n_0+1}|<1\Rightarrow|a_{n}|<|a_{n_0+1}\left|+1,\forall n>n_0\right.\)

Then the tail of \(a_n\) for \(n>n_0\) is bounded.

Let \(M=MAX\{|a_1|,|a_2|,...,|a_{n_0}|,|a_{n_0+1}|\}+1\Rightarrow\left|a_{n}\right|<M,\forall n\in\mathbb{N}\)

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\(\Leftarrow\)) Let \(a_n\) be a Cauchy sequence. Since the Lemma, it is bounded.

Then since Every bounded sequence (a_n) has a convergent subsequence, we have a convergent subsequence \(a_{n_k}\).

Let \(l=\lim_{k\to\infty}a_{n_{k}}\). Let's prove that \(l=\lim_{n\to\infty}a_{n}\)

Let \(\varepsilon>0\). We need \(n_0\) such that \(n>n_0\Rightarrow|a_{n}-l|<\varepsilon\)

Since \(l=\lim_{k\to\infty}a_{n_{k}}\), then \(\exists k_0\) such that \(k>k_0\Rightarrow|a_{n_k}-l|<\frac{\varepsilon}{2}\)

Since \(a_n\) is a Cauchy sequence, then \(\exists\tilde{n}_0\) such that \(n_1,n_2>\tilde{n}_0\Rightarrow|a_{n_1}-a_{n_2}|<\frac{\varepsilon}{2}\)

Since the definition of subsequence, \(\exists \tilde{k}_0\) such that \(k>\tilde{k}_0\Rightarrow n_k>n_0\)

Let \(n_0=\tilde{n}_0\). Pick \(k_1>k_0,\tilde{k}_0\Rightarrow\begin{cases}|a_{n_{k_1}}-l|<\frac{\varepsilon}{2}\\ n_{k_1}>\tilde{n}_0=n_0\end{cases}\)

Suppose \(n>n_0\Rightarrow|a_{n}-a_{n_{k_1}}|<\frac{\varepsilon}{2}\) (Since, let \(n_1=n,n_2=n_{k_1}\))

\(|a_{n}-l|=\left|a_{n}-a_{n_{k_1}}+a_{n_{k_1}}-l\right|\leq\left|a_{n}-a_{n_{k_1}}|+|a_{n_{k_1}}-l\right|<\varepsilon\)

Series

Definition

Given a sequence \((a_n)=(a_1,a_2,...)\), we construct a new sequence \((s_n)=(a_1,a_1+a_2,a_1+a_2+a_3,...)\)

\(S\) is the sum of the sequence \(a_n\)

Note

\((a_{n})_{n=1}^{\infty}=(a_{i})_{i=1}^{\infty}=(a_{x})_{x=1}^{\infty}\)

\(S_{n}=a_1+a_2+...+a_{n}=\sum_{i=1}^{n}a_{i=}\sum_{k=1}^{n}a_{k}\)

If \((a_{n})_{n=0}^{\infty},(S_{n})_{n=0}^{\infty},S_{n}=\sum_{k=0}^{n}a_{k}\)

Example

\(a_{n}=c,\forall n\geq1~~~~S_{n}=\sum_{k=1}^{n}a_{k}=c+c+c+\cdots+c\left(n~times\right)=nc\)
\(a_{n}=c,\forall n\geq0~~~~S_{n}=\sum_{k=0}^{n}a_{k}=c+c+c+\cdots+c\left(n+1~times\right)=\left(n+1\right)c\)
\(a_{n}=n,n\geq1~~~~S_{n}=\sum_{k=1}^{n}k=\frac{n\left(n+1\right)}{2}\)
\(a_{n}=n,\forall n\geq0~~~~S_{n}=\sum_{k=0}^{n}k=\frac{n\left(n+1\right)}{2}\)
\(a_{n}=\alpha^{n},0<\alpha<1,n\geq0\)

\(S_{n}=1+\alpha+\alpha^2+\alpha^3+\cdots+\alpha^{n}\)

\(\alpha S_{n}=\alpha+\alpha^2+\alpha^3+\alpha^4+\cdots+\alpha^{n+1}\)

\((1-\alpha)S_{n}=1-\alpha^{n+1}\Rightarrow S_{n}=\sum_{k=0}^{n}\alpha^{n}=\frac{1-\alpha^{n+1}}{1-\alpha}\) Geometric Series

\(a_{n}=\frac{1}{n},n\geq1~~~S_{n}=\sum_{k=1}^{n}\frac{1}{k}\) there is no nice formula Harmonic Series is divergent but harmonic sequence \(\left(\frac{1}{n}\right)_{n=1}^{\infty}\) is convergent

Series convergent

We say that a series \(\sum_{k=1}^{\infty}a_{k}\) is convergent if the sequence \((s_{n})=\sum_{k=1}^{n}a_{k}\) is convergent

In this case, let \(l=\lim_{n\to\infty}^{}S_{n}\), we denote \(\sum_{k=1}^{\infty}a_{k} \coloneqq l\)

That is by definition: \(\sum_{k=1}^{\infty}a_{k}=\lim_{n\to\infty}\sum_{k=1}^{n}a_{k}\)

Example

In previous examples

\(S_n=nc\) is not bounded \(\Rightarrow\) Not convergent
same
same
If \(\alpha >1\Rightarrow S_n\) is Not Bounded, not convergent

If \(0<\alpha<1\Rightarrow\alpha^{n+1}\rightarrow0,n\rightarrow\infty\), then \(S_{n}=\frac{1-\alpha^{n}}{1-\alpha}=\frac{1}{1-\alpha}-\frac{1}{1-\alpha}\alpha^{n+1}\xrightarrow{n\to\infty}\frac{1}{1-\alpha}\)

Therefore \(\sum_{k=0}^{\infty}\alpha^{k}=\frac{1}{1-\alpha}\)
\(\begin{aligned} 5) \quad S_n &= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} + \frac{1}{17} + \dots \end{aligned}\)

\(=1+\frac12+\left(\frac13+\frac14\right)+\left(\frac15+\frac16+\frac17+\frac18\right)+\left(\frac19+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)+\frac{1}{17}+\ldots\)

\(\geq1+\frac12+\left(\frac14+\frac14\right)+\left(\frac38+\frac18\right)+\left(\frac{8}{16}\right)+\cdots\)

\(\Rightarrow S_{2^{n}}^{}=1+\frac{n}{2}\quad\text{NOT BOUNDED, NOT convergent.}\)

Infinity

Given a sequence \((a_n)\), we say that \(\lim_{n \to \infty} a_n = \infty\) if \(\forall M>0,\exists n_0\) such that \(n \geq n_0\Rightarrow a_n>M\)

Theorem

If \((a_n)\) is monotonic increasing then \(\begin{cases}\text{if bounded ,}~~~~~\lim_{n\to\infty}a_{n}=sup\{a_{n}\}\\ \text{if not bounded,}\lim_{n\to\infty}a_{n}=\infty\end{cases}\)

Back to the examples

for \(c>0,\sum_{k=1}^{\infty}\alpha_{k}=\infty\)
\(\sum_{k=1}^{\infty}k=\infty\)
\(\alpha>1,\sum_{k=0}^{\infty}\alpha^{k}=\infty\)
\(\sum_{k=1}^{\infty}\frac{1}{k}=\infty\)

Remark

If \((a_{n})_{n=1}^{\infty}\) has the property \(a_{n}\geq0,\forall n\Rightarrow S_{n+1}=S_{n}+a_{n+1}\geq S_{n}\Rightarrow (S_n)\) is monotonic increasing

Thus \(\sum_{k=1}^{\infty}\alpha_{k}=\begin{cases}\text{if bounded, then convergnent to some }l\in\mathbb{R}_{\geq0}\\ \text{if not, then goes to }\infty\end{cases}\)

Conclusion: \(\sum_{k=1}^{\infty}\alpha_{k}=\begin{cases}l\\ \infty\end{cases}\)

For example

We saw \(\sum_{k=0}^{n}\alpha^{k}=\frac{1-\alpha^{n+1}}{1-\alpha}=\frac{\alpha^{n+1}-1}{\alpha-1}\) Geometric Series

If \(\alpha>1\Rightarrow\alpha^{n+1}\) is not bounded, then the series is \(+\infty\)
If \(0<\alpha<1\Rightarrow\alpha^{n+1}\xrightarrow{n\to\infty}0\), then the series is less than \(+\infty\) which is \(\frac{1}{1-\alpha}\)

In fact \(1\) is true if \(|\alpha|>1\) and \(2\) is true if \(|\alpha|<1\)