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10.28 Limit

Limit

Definition

Bound

\((a_n)\) is bounded if \(\exists M:|a_{n}|<M,\forall n\)

...bounded below if \(\exists M:a_{n}>M,\forall n\)

...bounded above if \(\exists M:a_{n}<M,\forall n\)

Important: look the tail of a sequence

Example

Suppose we know that \(|a_n|<100,\forall n>37000\), is \((a_n)\) bounded?

Suppose \(a_1=215,a_{39}=1357\), but it is bounded since \(|a_n|<MAX\{100,|a_1|,|a_2|,...,|a_{37000}|\}+1\)​

Proposition

If \((a_n)\) is convergent \(\Rightarrow (a_n)\) is bounded.

Proof

\(\begin{aligned}\text{Let }l=\lim_{n\to\infty}a_{n}\end{aligned}\)

Let \(\varepsilon=1\), then \(\exists n_0\) such that \(\forall n>n_0\) we have \(|a_n-l|<1\)

image​

Claim: \(|a_n|<|l|+1,\forall n>n_0\)

Analysis: We need \(|a_n|\leq M\) and we have \(|a_n-l|<1\), thus we want to use triangular inequality.

Trick: \(|a_{n}|=|a_{n}-l+l|\leq\left|a_{n}-l\right|+\left|l\right|<1+\left|l\right|=\left|l\right|+1\)

As in previous example: \(|a_n|<MAX\{|l|,|a_1|,|a_2|,...,|a_{n_0}|\}+1\)

The converse statement is false: A bounded sequence is convergent if FLASE

Example

  1. \(a_n=(-1)^n\), \(|a_n|<2,\forall n\) but not convergent

  2. \(a_{n}=\begin{cases}1~~~~~~~\text{ n is prime}\\ 0~~~~~~~\text{ n is composite}\end{cases}\)​

Monotonic

A sequence \((a_n)\) is

  • Monotonic increasing if \(a_{n+1}\geq a_n,\forall n\)​
  • Monotonic decreasing if \(a_{n+1}\leq a_{n},\forall n\)

Example

\(a_n=\frac1n\) is monotonic decreasing

Theorem

A is bounded monotonic increasing, then the sequence converges and \(\lim_{n\to\infty}a_{n}=sup\left(A\right)\)

Analysis

Since we need to prove \(\lim=\alpha\Rightarrow\left|a_{n}-\alpha\right|<\varepsilon\) and we know \(\alpha\) is a sup, then \(a_n\leq \alpha\) in particular

There is no absolute value of \(a_n\leq \alpha\)​\, thus we need to do some operation to remove the absolute value in \(|a_n-\alpha|<\varepsilon\)

Then we have \(\alpha-\varepsilon<a_n<\alpha+\varepsilon\) Obviously since the equivalence definition of supremum

Proof

Since \(a_n\) is bounded, then we can take sup

Let \(\alpha=sup(a_{n})\Rightarrow\begin{cases}a_{n}\leq\alpha,\forall n,\alpha\text{ is upper bound}~\textcircled{1}\\ \exists a_{n}:\alpha-\varepsilon<a_{n}<\alpha,\forall\varepsilon~\textcircled{2}\end{cases}\)

We want to prove \(\lim_{n\to\infty}a_{n}=\alpha\)

Pick \(\varepsilon>0\), we must find \(n_0:\forall n>n_0\Rightarrow\alpha-\varepsilon<a_{n}<\alpha+\varepsilon\)

Since \(\textcircled{1}\), then \(a_n\leq \alpha<\alpha+\varepsilon\)

Since \(\textcircled{2}\), \(\exists n_0:\alpha-\varepsilon<a_{n_0}\Rightarrow\forall n>n_0:a_{n}\geq a_{n_0}>\alpha-\varepsilon\) (monotonic)

Therefore, \(\forall n>n_0\Rightarrow\alpha-\varepsilon<a_{n}<\alpha+\varepsilon\)

Corollary

A is bounded monotonic decreasing, then the sequence converges and \(\lim_{n\to\infty}a_{n}=inf\left(A\right)\)

Proposition

A sequence that it tail \(\forall n>n_0\) is bounded and monotonic increasing \(\Rightarrow\) \((a_n)\) is convergent and \(\lim_{n\to\infty}a_{n}=sup\left(a_{n}\right)\left(n>n_0\right)\)

Assume it tail is bounded by \(M\)

Since its tail is bounded and the remaining element is finite, then \(|a_{n}|<MAX\left\lbrace|a_1|,|a_2|,...|a_{n_0}|,M\right\rbrace+1\)

Thus the sequence is bounded.

...Similarly

Algebra of limits

Consider \((a_n)_{n\in N}\),\((b_n)_{n\geq 1}\) two convergent sequences

\(\lim_{n\to\infty}a_{n}=a\) \(\lim_{n\to\infty}b_{n}=b\)

  1. \(\lim_{n\to\infty}(a_{n}+b_n)=a+b\)

    To prove this, let \(\varepsilon>0\). We need to find \(n_0\in N\)

    N.T.P \(|a_n+b_n-(a+b)|<\varepsilon\)

    Note that \(|a_{n}+b_{n}-(a+b)|\leq|a_{n}-a|+|b_{n}-b|\)

    Since \(a_n\rightarrow a\), we know that there is some \(n_0\in N||a_n-a|<\frac{\varepsilon}{2},~\forall n\geq n_0\)

    Similarly, \(b_n\rightarrow b\) so we can take \(n_1\in N||b_n-b|<\frac{\varepsilon}{2},~\forall n\geq n_1\) more strict restrict

    Then if \(n\geq max\{n_0,n_1\}\) we have that

    \(|a_n+b_n-(a+b)|\leq |a_n-a|+|n_b+b|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\)

  2. \(\lim_{n\to\infty}(a_{n}b_n)=ab\)

    To prove this, consider \(\varepsilon >0\)

    N.T.P. \(\exists n_0\in N||a_nb_n-ab|<\varepsilon~~\forall n\geq n_0\)

    Note that \(|a_nb_n-ab|=|a_nb_n+a_nb-a_nb-ab|\leq |a_n(b_n-b)|+|b||a_n-a|\leq |a_n||b_n-b|+|b||a_n-a|\)

    Now since \(a_n\) is convergent, there \(M\in R||a_n|\leq M~\forall n\in N\)

    Then \(|a_nb_n-ab|\leq M|b_n-b|+|b||a_n-a|\) spilit into two parts

    Consider \(n_1\in N||b_n-b|<\frac{\varepsilon}{2M}~\forall n\geq n_1(b_n\rightarrow b)\)

    Also, \(n_2\in N||a_n-a|<\frac{\varepsilon}{2|b|}~\forall n\geq n_2(a_n\rightarrow a)\)

    Then, if \(n\geq max\{n_1,n_2\},\) we have that

    \(|a_nb_n-ab|\leq M|b_n-b|+|b||a_n-a|<M\frac{\varepsilon}{2M}+|b|\frac{\varepsilon}{2|b|}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\)

    But \(|b|\) can be 0, thus if \(|b|=0\), then \(|a_{n}b_{n}-ab|\leq M|b_{n}-b|\leq M\frac{\varepsilon}{2M}=\frac{\varepsilon}{2}<\varepsilon\)​

  3. If \(a\neq 0\), \(\lim_{n\to\infty}\frac{1}{a_{n}}=\frac{1}{a}\)

    First, let \(\varepsilon=\frac{\left|a\right|}{2}>0\) because \(|a|\neq 0\)​

    Since \(\lim_{n\to\infty}a_{n}=a,\exists\tilde{n}:n>\tilde{n}\Rightarrow a-\frac{\left|a\right|}{2}<a_{n}<a+\frac{\left|a\right|}{2}\Rightarrow\begin{cases}if~a>0\Rightarrow0<a-\frac{a}{2}=\frac{a}{2}<a_{n}\\ if~a<0\Rightarrow a_{n}<a-\frac{a}{2}=\frac{a}{2}<0\end{cases}\Rightarrow a_{n}\neq0,\forall n>\tilde{n},\left|a_{n}\right|>\frac{\left|a\right|}{2}\)​

    image​

    We know \(\forall \varepsilon_1>0:\exists n_1:n>n_1\Rightarrow|a_n-a|<\varepsilon_1\)

    Given \(\varepsilon>0\) we want \(n_0\) such that \(n>n_0\Rightarrow\left|\frac{1}{a_{n}}-\frac{1}{a}\right|<\varepsilon\)​

    Let \(\varepsilon_1=\frac{\left|a\right|^2\varepsilon}{2}\Rightarrow n_1:n>n_1\Rightarrow\left|a_{n}-a\right|<\frac{\left|a\right|^2\varepsilon}{2}\)​

    Now let \(n_0=MAX\{n_{1,}\tilde{n}\}\)​

    If \(n>n_0\Rightarrow n>\tilde{n},n_1\Rightarrow\left|\frac{1}{a_{n}}-\frac{1}{a}\right|=|\frac{a-a_{n}}{a_{n}a}|=\frac{\left|a_{n}-a\right|}{\left|a_{n}\right|\left|a\right|}<\frac{\left|a_{n}-a\right|}{\frac{\left|a\right|}{2}|a|}=\frac{2\left|a_{n}-a\right|}{\left|a\right|^2}<\frac{2}{\left|a\right|^2}\varepsilon_1=\varepsilon\)​

  4. If \(b\neq 0\), \(\lim_{n\to\infty}\frac{a_{n}}{b_n}=\frac{a}{b}\)

    Lemma 1: If \(\lim_{n \to \infty} a_n = \ell\) and \(\ell > 0\), then there is a number \(X\) such that \(a_n > \frac{1}{2} \ell\) for all \(n > X\).

    Proof: Since \(\frac{1}{2} \ell > 0\), the terms \(a_n\) must eventually lie within a distance \(\frac{1}{2} \ell\) of the limit \(\ell\).

    In other words, there is a number \(X\) such that \(|a_n - \ell| < \frac{1}{2} \ell\) for all \(n > X\)

    Hence \(-\frac{1}{2} \ell < a_n - \ell < \frac{1}{2} \ell\), for all \(n > X\) and so the left-hand inequality gives

    \(\frac{1}{2} \ell < a_n\), for all \(n > X\), as required.

    Proof: We assume that \(m > 0\); the proof for the case \(m < 0\) is similar. Once again, the idea is to write the required expression in terms of \(a_n - \ell\) and \(b_n - m\):

    \[ \frac{a_n}{b_n} - \frac{\ell}{m} = \frac{m(a_n - \ell) - \ell(b_n - m)}{b_n m}. \]

    Now, however, there is a slight problem: \(\{m(a_n - \ell) - \ell(b_n - m)\}\) is certainly a null sequence, but the denominator is rather awkward. Some of the terms \(b_n\) may take the value \(0\), in which case the expression is undefined.

    However, by Lemma 1, we know that for some \(X\) we have \(b_n > \frac{1}{2} m, \quad \text{for all } n > X.\)​

    Thus, for all \(n > X\):

    \[ \left|\frac{a_n}{b_n} - \frac{\ell}{m}\right| = \frac{|m(a_n - \ell) - \ell(b_n - m)|}{b_n m}\leq \frac{|m(a_n - \ell) - \ell(b_n - m)|}{\frac{1}{2} m^2}\leq \frac{|m| \times |a_n - \ell| + |\ell| \times |b_n - m|}{\frac{1}{2} m^2}. \]

Example

  1. We know \(\lim_{n\to\infty}\frac{1}{n}=0\Rightarrow\lim_{n\to\infty}\frac{1}{n^2}=0\) follows from theorem 2 \(\Rightarrow\lim_{n\to\infty}\frac{1}{n^{k}}=0,\forall k>0\) induction

  2. If \(a_{n}=c\in\mathbb{R},\forall n\Rightarrow\lim_{n\to\infty}a_{n}=c\)​

    Given \(\varepsilon>0\), take \(n_0=1\), \(\forall n>n_0,|a_n-c|<\varepsilon\)​

  3. \(\lim_{n\to\infty}\frac{2}{n}=0\) follows from theorem 1 or follows from theorem 2 (using example 2)

  4. \(\lim_{n\to\infty}\frac{3n+\pi+\frac{1}{\pi}n^2}{n^2}=\lim_{n\to\infty}\frac{3n}{n^2}+\lim_{n\to\infty}\frac{\pi}{n^2}+\lim_{n\to\infty}\frac{\frac{1}{\pi}n^2}{n^2}=\frac{1}{\pi}\)​

Lemma (Sandwich)

Suppose \(a_n\leq b_n\leq c_n\) and \(\lim_{n\to\infty}a_{n}=\lim_{n\to\infty}c_{n}=l\Rightarrow\lim_{n\to\infty}b_{n}=l\)

Proof

We have \(l-\varepsilon<a_n<l+\varepsilon\) and \(l-\varepsilon<c_n<l+\varepsilon\), then \(l-\varepsilon<a_n\) and \(c_n<l+\varepsilon\)

Then since \(a_n\leq b_n\leq c_n\), then \(l-\varepsilon<b_n<l+\varepsilon\)

Easily, \(\lim b_n=l\)​

Example

Let \(0<\alpha<1\), prove \(\lim_{n\to\infty}\alpha^{n}=0\)​

Let us do the case \(\alpha=\frac12\), we have \(0<\left(\frac12\right)^{n}<\frac{1}{n}\)

Since \(\lim_{n\to\infty}\frac{1}{n}=0\), then we can apply it.

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Exercise: do it for arbitrary \(\alpha\)

  1. If \(\alpha >1\Rightarrow a_n=\alpha^n\) is NOT Bounded

    Suppose that \(\{\alpha^n\}\) is bounded above, then let \(s=sup\{\alpha^{n}\}\)​

    Then \(s\geq\alpha^{n},\forall n\), then \(s\geq\alpha^{n+1}\Rightarrow\frac{s}{\alpha}\geq\alpha^{n},\forall n\)​

    Since \(\alpha>1\), then \(\frac{s}{\alpha}<s\), then \(\frac{s}{\alpha}\) is a supremum

    Contradiction!

  2. If \(0\leq \alpha <1\Rightarrow a_n=\alpha^n\) is convergent, \(\lim_{n\to\infty}a_{n}=0\)​

    Two proofs

    1. If \(\alpha=0\Rightarrow a_{n}=0,\forall n\Rightarrow\lim_{n\to\infty}a_{n}=0\)​

      If \(0<\alpha<1\Rightarrow\frac{1}{\alpha}>1\Rightarrow\left(\frac{1}{\alpha}\right)^{n}\) is not bounded above since 1

      Given \(\varepsilon>0\), let \(n_0\) such that \((\frac{1}{\alpha})^{n_0}>\frac{1}{\varepsilon}\Rightarrow\forall n>n_0:(\frac{1}{\alpha})^{n}>(\frac{1}{\alpha})^{n_0}>\frac{1}{\varepsilon}\Rightarrow\frac{1}{\alpha^{n}}>\frac{1}{\varepsilon}\Rightarrow\varepsilon>\alpha^{n}\Rightarrow\left|\alpha^{n}-0\right|<\varepsilon\)​ 2. \(a_n=\alpha^n\)

      Since \(0<\alpha<1\Rightarrow\alpha^{n+1}<\alpha^{n}\Rightarrow a_{n+1}<a_{n}\Rightarrow\) monotonic decreasing

      And \(\{a_n\}\) is bounded below by \(0\)​

      Then \(a_n\) is convergent to the \(infimum=l\)

      Assume \(\lim_{n\to\infty}a_{n}=l\Rightarrow\lim_{n\to\infty}a_{n+1}=l\)​

      Since \(a_{n+1}=\alpha\cdot a_{n}\Rightarrow\lim_{n\to\infty}a_{n}=\alpha\cdot\lim_{n\to\infty}a_{n+1}\Rightarrow l=\alpha\cdot l\)​

      Since \(\alpha\neq 1\), then \(l=0\)

Method

3 Method to cope with the limit:

  1. definition: directly take \(\varepsilon\) or the expression of \(\varepsilon\)

  2. definition but take \(\varepsilon>\frac1n\) ... . You should prove \(|a_n-L|\) is bounded by \(\frac1n\)

    (20 points) Prove that for all real number L \in (0,1), exists a sequence of rational numbers (a... use density to prove bounded by \(\frac1n\) here

    Let A be a non-empty bounded set and s=\sup A. Prove that there exists a sequence (a_n) such tha... use the property of sup to prove bounded by \(\frac1n\) here

    only when the question is about proving the existence of sequence!!!

  3. take subsequence and use the limit of subsequence equals to limit of sequence

    The sequence must have a specific expression firstly.

  4. Squeeze theorem

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