# 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. ${\displaystyle \lim _{n\to \infty }{\frac {\sin {\mathrm {e} ^{n}}}{n}}=0}$

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

quod erat demonstrandum

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

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

quod erat demonstrandum

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

Proof. Divide both numerator and denominator by ${\displaystyle n^{2}}$:

{\displaystyle {\begin{aligned}{\frac {(1+2n)^{2}}{1+2n^{2}}}&={\frac {\frac {(1+2n)^{2}}{n^{2}}}{\frac {1+2n^{2}}{n^{2}}}}={\frac {\left({\frac {1}{n}}+2\right)^{2}}{\left({\frac {1}{n}}\right)^{2}+2}}\end{aligned}}}

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

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

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

quod erat demonstrandum

## Monotone Sequences

A sequence ${\displaystyle \left\{x_{n}\right\}}$ is called increasing (or nondecreasing) if ${\displaystyle x_{1}\leq x_{2}\leq x_{3}\leq \dots }$, or, to be precise, ${\displaystyle x_{n}\leq x_{n+1}}$ for all ${\displaystyle n\in \mathbb {N} }$.

It is called strictly increasing if ${\displaystyle x_{n} for all ${\displaystyle n\in \mathbb {N} }$.

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

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

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

### Examples

• ${\displaystyle \left\{{\frac {1}{n}}\right\}}$ 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 ${\displaystyle a}$, ${\displaystyle a}$, … is both increasing and decreasing

### Theorems

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

Proof. Let ${\displaystyle \left\{x_{n}\right\}}$ be an increasing sequence.

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

We claim that ${\displaystyle x_{n}\to \sup {E}}$ as ${\displaystyle n\to \infty }$. Take any ${\displaystyle \epsilon >0}$. Then ${\displaystyle \sup {E}-\epsilon }$ is not an upper bound of ${\displaystyle E}$. Hence there exists ${\displaystyle n_{0}\in \mathbb {N} }$ such that ${\displaystyle x_{n_{0}}>\sup {E}-\epsilon }$. Since the sequence is increasing, it follows that ${\displaystyle x_{n}\geq x_{n_{0}}>\sup {E}-\epsilon }$ for ${\displaystyle n\geq n_{0}}$. At the same time, ${\displaystyle x_{n}\leq \sup {E}}$ for all ${\displaystyle n\in \mathbb {N} }$. Therefore ${\displaystyle \left|x_{n}-\sup {E}\right|<\epsilon }$ for all ${\displaystyle n\geq n_{0}}$, which proves the claim.

Now consider the case when ${\displaystyle \left\{x_{n}\right\}}$ is not bounded. Note that the set ${\displaystyle E}$ is bounded below (as ${\displaystyle x_{1}}$ is a lower bound). Hence ${\displaystyle E}$ is not bounded above. Then for any ${\displaystyle C\in \mathbb {R} }$, there exists ${\displaystyle n_{0}\in \mathbb {N} }$ such that ${\displaystyle x_{n_{0}}>C}$. It follows that ${\displaystyle x_{n}\geq x_{n_{0}}>C}$ for all ${\displaystyle n\geq n_{0}}$. Thus ${\displaystyle \left\{x_{n}\right\}}$ diverges to ${\displaystyle +\infty }$.

quod erat demonstrandum

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

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

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 ${\displaystyle I_{1},I_{2},\ldots }$ is called nested if ${\displaystyle I_{1}\supset I_{2}\supset \ldots }$. That is, ${\displaystyle I_{n}\supset I_{n+1}}$ for all ${\displaystyle n\in \mathbb {N} }$.

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

Proof. Let ${\displaystyle I_{n}=\left[a_{n},b_{n}\right]}$ for ${\displaystyle n\in \mathbb {N} }$. Since the sequence ${\displaystyle \left\{I_{n}\right\}}$ is nested, it follows that the sequence ${\displaystyle \left\{a_{n}\right\}}$ is increasing while the sequence ${\displaystyle \left\{b_{n}\right\}}$ is decreasing. Besides, both sequences are bonded (since both are contained in the bounded interval ${\displaystyle I_{1}}$). Hence both are convergent: ${\displaystyle a_{n}\to a}$ and ${\displaystyle b_{n}\to b}$ as ${\displaystyle n\to \infty }$. Since ${\displaystyle a_{n}\leq b_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$, the Comparison Theorem implies that ${\displaystyle a\leq b}$.

We claim that ${\displaystyle \bigcap _{n\in \mathbb {N} }I_{n}=\left[a,b\right]}$. Indeed, we have ${\displaystyle a_{n}\leq a}$ for all ${\displaystyle n\in \mathbb {N} }$ (by Comparison Theorem applied to ${\displaystyle a_{1},a_{2},\ldots }$ and constant sequence ${\displaystyle a_{n},a_{n},\ldots }$). Similarly, ${\displaystyle b\leq b_{n}}$ for all ${\displaystyle n\in \mathbb {N} }$. Therefore ${\displaystyle \left[a,b\right]}$ is contained in the intersection.

On the other hand, if ${\displaystyle x, then ${\displaystyle x for some ${\displaystyle n}$ so that ${\displaystyle x\not \in I_{n}}$. Similarly, if ${\displaystyle x>b}$ then ${\displaystyle x>b_{m}}$ for some ${\displaystyle m}$ so that ${\displaystyle x\not \in I_{m}}$. This proves the claim.

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

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A few remarks:

1. The theorem may not hold if the intervals are open. Counterexample: ${\displaystyle I_{n}=\left(0,{\frac {1}{n}}\right)}$, ${\displaystyle n\in \mathbb {N} }$ The intervals are nested and bounded, but their intersection is empty since ${\displaystyle {\frac {1}{n}}\to 0}$ as ${\displaystyle n\to \infty }$.
2. The theorem may not hold if the intervals are not bounded. Counterexample: ${\displaystyle I_{n}=\left[n,\infty \right)}$, ${\displaystyle n\in \mathbb {N} }$. The intervals are nested and closed, but their intersection is empty since the sequence ${\displaystyle \left\{n\right\}}$ diverges to ${\displaystyle +\infty }$.
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 ${\displaystyle \left\{x_{n}\right\}}$ be a bounded sequence of real numbers. We are going to build a nested sequence of intervals ${\displaystyle I_{n}=\left[a_{n},b_{n}\right]}$ for ${\displaystyle n\in \mathbb {N} }$, such that each ${\displaystyle I_{n}}$ contains infinitely many elements of ${\displaystyle \left\{x_{n}\right\}}$ and ${\displaystyle \left|I_{n+1}\right|={\frac {\left|I_{n}\right|}{2}}}$ for all ${\displaystyle n\in \mathbb {N} }$. The sequence is built inductively.

Basis. First we set ${\displaystyle I_{1}}$ to be any closed bounded interval that contains all elements of ${\displaystyle \left\{x_{n}\right\}}$ (such an interval exists because the sequence ${\displaystyle \left\{x_{n}\right\}}$ is bounded).

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

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

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

Basis. First let ${\displaystyle n_{1}=1}$.

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

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

quod erat demonstrandum