MATH 409 Lecture 7

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Lecture Slides

Review

• Limits of Sequences
• Properties of Convergent sequences
  • Limit is unique
  • Convergent sequence is bounded
  • Subsequence of convergent sequence converges to same ilmit
  • Modifying a finite number of elements can't affect convergence or change its limit
  • Rearranging elements of a sequence can't affect convergence or change its limit
• Limit Theorems

Examples

Theorem. $\underset{n\to \mathrm{\infty }}{\lim }\frac{\sin {\mathrm{e}}^n}{n}=0$

Proof. $-\frac{1}{n}\le \frac{\sin {\mathrm{e}}^n}{n}\le \frac{1}{n}$ for all $n\in \mathbb{\mathbb{N}}$ since $-1\le \sin x\le 1$ for all $x\in \mathbb{ℝ}$. As shown last time, $\frac{1}{n}\to 0$ as $n\to \mathrm{\infty }$. Then $-\frac{1}{n}\to -1\cdot 0=0$ as $n\to \mathrm{\infty }$. By the Squeeze theorem, $\frac{\sin {\mathrm{e}}^n}{n}=0$ as $n\to \mathrm{\infty }$.

quod erat demonstrandum

Theorem. $\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{2^n}=0$

Proof. The sequence $\left\{\frac{1}{2^n}\right\}$ is a subsequence of $\left\{\frac{1}{n}\right\}$. Hence it converges to the same limit.

quod erat demonstrandum

Theoreom. $\underset{n\to \mathrm{\infty }}{\lim }\frac{(1+2n{)}^2}{1+2n^2}=2$

Proof. Divide both numerator and denominator by $n^2$:

$\begin{array}{rl}\frac{(1+2n{)}^2}{1+2n^2}& =\frac{\frac{(1+2n{)}^2}{n^2}}{\frac{1+2n^2}{n^2}}=\frac{{(\frac{1}{n}+2)}^2}{{\left(\frac{1}{n}\right)}^2+2}\end{array}$

Since $\frac{1}{n}\to 0$ as $n\to \mathrm{\infty }$, it follows that

• $\frac{1}{n}+2\to 0+2=2$ as $n\to \mathrm{\infty }$,
• ${(\frac{1}{n}+2)}^2\to 2^2=4$ as $n\to \mathrm{\infty }$
• ${\left(\frac{1}{n}\right)}^2\to 0^2=0$ as $n\to \mathrm{\infty }$
• ${\left(\frac{1}{n}\right)}^2+2\to 0+2=2$ as $n\to \mathrm{\infty }$

Finally, $\frac{{(\frac{1}{n}+2)}^2}{{\left(\frac{1}{n}\right)}^2+2}\to \frac{4}{2}=2$ as $n\to \mathrm{\infty }$.

quod erat demonstrandum

Monotone Sequences

A sequence $\left\{x_n\right\}$ is called increasing (or nondecreasing) if $x_1\le x_2\le x_3\le \dots$, or, to be precise, $x_n\le x_{n+1}$ for all $n\in \mathbb{\mathbb{N}}$.

It is called strictly increasing if $x_n<x_{n+1}$ for all $n\in \mathbb{\mathbb{N}}$.

Note: These definitions are according to the textbooks. Other people prefer to use nondecreasing for ≤ and increasing for strictly increasing.

Similarly, $\left\{x_n\right\}$ is called decreasing (or nonincreasing) if $x_n\ge x_{n+1}$ for all $n\in \mathbb{\mathbb{N}}$.

It is called strictly decreasing if $x_n>x_{n+1}$ for all $n\in \mathbb{\mathbb{N}}$.

Examples

• $\left\{\frac{1}{n}\right\}$ is strictly decreasing
• The sequence 1, 1, 2, 2, 3, 3, … is increasing, but not strictly increasing
• The sequence -1, 1, -1, 1, … is neither increasing nor decreasing
• A constant sequence $a$, $a$, … is both increasing and decreasing

Theorems

Theorem. Any increasing sequence converges to a limit if it is bounded, and diverges to $+\mathrm{\infty }$ otherwise.

Proof. Let $\left\{x_n\right\}$ be an increasing sequence.

First consider the case when $\left\{x_n\right\}$ is bounded. In this case, the set $E$ of all elements occurring in the sequence is bounded. By the completeless axiom, any bounded set has a supremum; thus $\sup E$ exists.

We claim that $x_n\to \sup E$ as $n\to \mathrm{\infty }$. Take any $ϵ>0$. Then $\sup E-ϵ$ is not an upper bound of $E$. Hence there exists $n_0\in \mathbb{\mathbb{N}}$ such that $x_{n_0}>\sup E-ϵ$. Since the sequence is increasing, it follows that $x_n\ge x_{n_0}>\sup E-ϵ$ for $n\ge n_0$. At the same time, $x_n\le \sup E$ for all $n\in \mathbb{\mathbb{N}}$. Therefore $|x_n-\sup E|<ϵ$ for all $n\ge n_0$, which proves the claim.

Now consider the case when $\left\{x_n\right\}$ is not bounded. Note that the set $E$ is bounded below (as $x_1$ is a lower bound). Hence $E$ is not bounded above. Then for any $C\in \mathbb{ℝ}$, there exists $n_0\in \mathbb{\mathbb{N}}$ such that $x_{n_0}>C$. It follows that $x_n\ge x_{n_0}>C$ for all $n\ge n_0$. Thus $\left\{x_n\right\}$ diverges to $+\mathrm{\infty }$.

quod erat demonstrandum

Theorem. Any decreasing sequence converges to a limit if it is bounded, and diverges to $-\mathrm{\infty }$ otherwise.

Proof. Let $\left\{x_n\right\}$ be a decreasing sequence. Then the sequence $\{-x_n\}$ is increasing since $a\ge b$ is equivalent to $-a\le -b$ for all $a,b\in \mathbb{ℝ}$. By the previous theorem, either $-x_n\to c$ for some $c\in \mathbb{ℝ}$ as $n\to \mathrm{\infty }$, or else $-x_n$ diverges to $+\mathrm{\infty }$. In the former case, $x_n\to -c$ as $n\to \mathrm{\infty }$. In the latter case, $x_n$ diverges to $-\mathrm{\infty }$.

quod erat demonstrandum

Corollary. Any monotone sequence converges to a limit if it is bounded, and diverges to infinity otherwise.

Nested Intervals Property

A sequence of sets $I_1,I_2,\dots$ is called nested if $I_1\supset I_2\supset \dots$. That is, $I_n\supset I_{n+1}$ for all $n\in \mathbb{\mathbb{N}}$.

Theorem. If $\left\{I_n\right\}$ is a nested sequence of nonempty closed bounded intervals, then the intersection $\bigcap _{n\in \mathbb{\mathbb{N}}}{I}_n$ is nonempty. Moreover, if lengths $\left|I_n\right|$ of the intervals satisfy $\left|I_n\right|\to 0$ as $n\to \mathrm{\infty }$, then the intersetion consists of a single point.

Proof. Let $I_n=[a_n,b_n]$ for $n\in \mathbb{\mathbb{N}}$. Since the sequence $\left\{I_n\right\}$ is nested, it follows that the sequence $\left\{a_n\right\}$ is increasing while the sequence $\left\{b_n\right\}$ is decreasing. Besides, both sequences are bonded (since both are contained in the bounded interval $I_1$). Hence both are convergent: $a_n\to a$ and $b_n\to b$ as $n\to \mathrm{\infty }$. Since $a_n\le b_n$ for all $n\in \mathbb{\mathbb{N}}$, the Comparison Theorem implies that $a\le b$.

We claim that $\bigcap _{n\in \mathbb{\mathbb{N}}}{I}_n=[a,b]$. Indeed, we have $a_n\le a$ for all $n\in \mathbb{\mathbb{N}}$ (by Comparison Theorem applied to $a_1,a_2,\dots$ and constant sequence $a_n,a_n,\dots$). Similarly, $b\le b_n$ for all $n\in \mathbb{\mathbb{N}}$. Therefore $[a,b]$ is contained in the intersection.

On the other hand, if $x<a$, then $x<a_n$ for some $n$ so that $x\in ̸I_n$. Similarly, if $x>b$ then $x>b_m$ for some $m$ so that $x\in ̸I_m$. This proves the claim.

Clearly, the length of $[a,b]$ cannot exceed $\left|I_n\right|$ for any $n\in \mathbb{\mathbb{N}}$. Therefore $\left|I_n\right|\to 0$ as $n\to \mathrm{\infty }$ implies that $[a,b]$ is a degenerate interval: $a=b$.

quod erat demonstrandum

A few remarks:

1. The theorem may not hold if the intervals are open. Counterexample: $I_n=(0,\frac{1}{n})$, $n\in \mathbb{\mathbb{N}}$ The intervals are nested and bounded, but their intersection is empty since $\frac{1}{n}\to 0$ as $n\to \mathrm{\infty }$.
2. The theorem may not hold if the intervals are not bounded. Counterexample: $I_n=[n,\mathrm{\infty })$, $n\in \mathbb{\mathbb{N}}$. The intervals are nested and closed, but their intersection is empty since the sequence $\left\{n\right\}$ diverges to $+\mathrm{\infty }$.
3. This is another form of the completeness axiom.
4. This theorem can be used to prove the existence of a decimal expansion

Bolzano-Weierstrass Theorem

Theorem. Every bounded sequence of real numbers has a convergent subsequence.

Proof. Let $\left\{x_n\right\}$ be a bounded sequence of real numbers. We are going to build a nested sequence of intervals $I_n=[a_n,b_n]$ for $n\in \mathbb{\mathbb{N}}$, such that each $I_n$ contains infinitely many elements of $\left\{x_n\right\}$ and $\left|I_{n+1}\right|=\frac{\left|I_n\right|}{2}$ for all $n\in \mathbb{\mathbb{N}}$. The sequence is built inductively.

Basis. First we set $I_1$ to be any closed bounded interval that contains all elements of $\left\{x_n\right\}$ (such an interval exists because the sequence $\left\{x_n\right\}$ is bounded).

Induction. Now assume that for some $n\in \mathbb{\mathbb{N}}$ the interval $I_n$ is already chosen and it contains infinitely many elements of $\left\{x_n\right\}$. Then at least one of the subintervals $I^{\prime }=[a_n,\frac{a_n+b_n}{2}]$ and $I^{″}=[\frac{a_n+b_n}{2},b_n]$ also contains infinitely many elements of $\left\{x_n\right\}$. We set $I_{n+1}$ to be such an interval. By construction, $I_{n+1}\subset I_n$ and $\left|I_{n+1}\right|=\frac{\left|I_n\right|}{2}$.

Since $\left|I_{n+1}\right|=\frac{\left|I_n\right|}{2}$ for all $n\in \mathbb{\mathbb{N}}$, it follows by induction that $\left|I_n\right|=\frac{\left|I_1\right|}{2^{n-1}}$ for all $n\in \mathbb{\mathbb{N}}$. As a consequence, $\left|I_n\right|\to 0$ as $n\to \mathrm{\infty }$. By the Nested Intervals Property, the intersection of the intervals consists of a single number $a$.

Next we are going to build a strictly increasing sequence of natural numbers $\left\{n_k\right\}$ such that $x_{n_k}\in I_k$ for all $k\in \mathbb{\mathbb{N}}$. The sequence is built inductively:

Basis. First let $n_1=1$.

Induction. Now assume that for some $k\in \mathbb{\mathbb{N}}$ the number $n_k$ is already chosen. Since the interval $I_{k+1}$ contains infinitely many elements of the sequence $\left\{x_n\right\}$, there exists $m>n_k$ such that $x_m\in I_{k+1}$. We set $n_{k+1}=m$.

Now we claim that the subsequence ${\left\{x_{n_k}\right\}}_{k\in \mathbb{\mathbb{N}}}$ of the sequence $\left\{x_n\right\}$ converges to $a$. Indeed, for any $k\in \mathbb{\mathbb{N}}$, the points $x_{n_k}$ and $a$ both belong to the interval $I_k$. Hence $|x_{n_k}-a|\le \left|I_k\right|$. Since $\left|I_k\right|\to 0$ as $k\to \mathrm{\infty }$, it follows that $x_{n_k}\to a$ as $k\to \mathrm{\infty }$.

quod erat demonstrandum