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10.24 Convergence

Sequences

Definition

A sequence is a function \(a:\N\rightarrow \R\)

Notation

\(a_n=a(n)\)

Say: \((a_n)_{n\in\N}\) is a sequence (\(a_1,a_2,a_3,....\))

Example

\(a_n=2n+1,(a_n)=(3,5,7,...)\)

Constant Sequence: \(a_n=1,\forall n\)

\(f_{n}=\begin{cases}1\text{, if }n=1,2\\f_{n-1}+f_{n-2},n\geq 3\end{cases}\)

\(\begin{aligned}a_{n}=1+2+...+n=\sum_{i=1}^{n}i\end{aligned}\) \(a=(1,3,6,10,...)\)

\(\begin{aligned}a_{n}=1-2+3-4+5-6+...\pm n\end{aligned}\) \(a=(1,-1,2,-2,...)\)

\(\begin{aligned}a_{n}=\frac11+\frac12+\frac13+...+\frac{1}{n}=\sum_{i=1}^{n}\frac{1}{i}\end{aligned}\) \(a=(1,\frac32,\frac{11}{6},...)\)​

\(\begin{aligned}a_{n}=-\frac11+\frac12-\frac13+...\pm\frac{1}{n}=\sum_{i=1}^{n}\frac{\left(-1\right)^{i}}{i}\end{aligned}\)

Limit of sequence

Definition

Convergence

We say that a sequence \(a_n\) is convergent if there is a number \(l\in\mathbb{R}\) such that \(\forall\varepsilon>0,\exists n_0\in\mathbb{N}\) with the property

\(\begin{cases}\forall n>n_0,l-\varepsilon<a_{n}<l+\varepsilon\\ \forall n>n_0,\left|a_{n}-l\right|<\varepsilon\\ \forall n>n_0,-\varepsilon<a_{n}-l<\varepsilon\end{cases}\), this three properties are equivalent.

Denotation

\(\lim_{n\rightarrow\infty}a_{n}=l\) or \(a_n\xrightarrow{n\to\infty}l\)​

Divergence

If the previous does not happen, we say \(a_n\) is divergent.

Example

  1. \(\begin{aligned}\lim_{n\rightarrow\infty}\frac{1}{n}=0\end{aligned}\)

    Proof

    Let \(\varepsilon>0\), we need to find \(n_0\in \N\) such that \(|\frac1n-0|<\varepsilon,\forall n>n_0\)​

    Let \(n_0\) be a natural number such that \(n_0>\frac{1}{\varepsilon}\) (Archimedes property)

    Now, \(\forall n>n_0\) we have \(n>n_0>\frac{1}{\varepsilon}\Rightarrow n>\frac{1}{\varepsilon}\Rightarrow\varepsilon>\frac{1}{n}\Rightarrow\frac{1}{n}<\varepsilon\Rightarrow\left|\frac{1}{n}-0\right|<\varepsilon\)​

  2. \(\begin{aligned}\lim_{n\rightarrow\infty}\frac{n^2+1}{n^2}=1\end{aligned}\) (\(\equiv1+\frac{1}{n^2}\))

    Proof

    Let \(\varepsilon>0\), we need to find \(n_0\in \N\) such that \(|\frac{n^2+1}{n^2}-1|<\varepsilon,\forall n>n_0\)

    Let \(n_0\) be a natural number such that \(n_0>\frac{1}{\varepsilon}\) (Archimedes property)

    Now, \(\forall n>n_0\) we have \(n^2>n>n_0>\frac{1}{\varepsilon}\Rightarrow n^2>\frac{1}{\varepsilon}\Rightarrow\varepsilon>\frac{1}{n^2}\Rightarrow\frac{1}{n^2}<\varepsilon\Rightarrow|\frac{1}{n^2}|<\varepsilon\Rightarrow|\frac{n^2+1}{n^2}-1|<\varepsilon\)

    (Note: We can choose \(n^2>\frac{1}{\varepsilon}\Rightarrow n>\frac{1}{\sqrt{\varepsilon}}\), this is very precise and the above method is more bold and clumsy and completely enough)

Theorem

If \(\begin{aligned}\lim_{n\rightarrow\infty}a_{n}=l_1\end{aligned}\) and \(\begin{aligned}\lim_{n\rightarrow\infty}a_{n}=l_2\end{aligned}\), then \(l_1=l_2\)

Analysis: How to prove the limit is unique? We can image if their limit are different, then "the window\" must at least have the no same part or totally different. The easiest way is to let the line overlap because it is easy to take the epsilon since you don't know the specific sequence and range.

Proof

Suppose \(l_1<l_2\) and let \(\varepsilon=\frac12\left(l_2-l_1\right)>0\)

image​

By definition: \(\exists n_1\in\mathbb{N}\) such that \(\forall n>n_1\Rightarrow a_{n}\in\left(l_1-\varepsilon,l_1+\varepsilon\right)\)

\(\exists n_2\in\mathbb{N}\) such that \(\forall n>n_2\Rightarrow a_{n}\in\left(l_2-\varepsilon,l_2+\varepsilon\right)\)

Then \(a_n<l_1+\varepsilon\) and \(a_n>l_2-\varepsilon\) \(\Rightarrow a_{n}<\frac12\left(l_1+l_2\right)\) and \(a_{n}>\frac12\left(l_1+l_2\right)\) ,\(\forall n>\max\left(n_1,n_2\right)\) which is a contradiction!

Remark

We can think in a set associated to a sequence \((a_{n})=(a_1,a_2,a_3,\ldots)\)

The set: \(\{a_1,a_2,a_3,...\}\) (without reputation 集合里面不能有重复)

More definitions

Given a sequence (\(a_n\)), we say that the sequence is bounded above or bounded below if so is the set \(\{a_n\}\)

The sequence is bounded if it is bounded above and bounded below

Given a sequence (\(a_n\)), an upper bounds of \(a_n\) is an upper bound of the set \(\{a_n\}\)

Given a sequence (\(a_n\)), an lower bounds of \(a_n\) is an lower bound of the set \(\{a_n\}\)

Given a sequence (\(a_n\)), an maximum of \(a_n\) is an maximum of the set \(\{a_n\}\)

Given a sequence (\(a_n\)), an minimum of \(a_n\) is an minimum of the set \(\{a_n\}\)

Given a sequence (\(a_n\)), a supremum of \(a_n\) is a supremum of the set \(\{a_n\}\)

Given a sequence (\(a_n\)), an infimum of \(a_n\) is an infimum of the set \(\{a_n\}\)

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