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

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

quod erat demonstrandum

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

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

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

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

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

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

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

quod erat demonstrandum

Monotone Sequences

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

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

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

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

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

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

quod erat demonstrandum

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

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

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

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

Proof. Let $I_{n}=\left[a_{n},b_{n}\right]$ for $n\in \mathbb {N}$. 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 $n\to \infty$. Since $a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$, the Comparison Theorem implies that $a\leq b$.

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

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

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

quod erat demonstrandum

A few remarks:

1. The theorem may not hold if the intervals are open. Counterexample: $I_{n}=\left(0,{\frac {1}{n}}\right)$, $n\in \mathbb {N}$ The intervals are nested and bounded, but their intersection is empty since ${\frac {1}{n}}\to 0$ as $n\to \infty$.
2. The theorem may not hold if the intervals are not bounded. Counterexample: $I_{n}=\left[n,\infty \right)$, $n\in \mathbb {N}$. The intervals are nested and closed, but their intersection is empty since the sequence $\left\{n\right\}$ diverges to $+\infty$.
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 $I_{n}=\left[a_{n},b_{n}\right]$ for $n\in \mathbb {N}$, 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 $n\in \mathbb {N}$. 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 $n\in \mathbb {N}$ 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 $I'=\left[a_{n},{\frac {a_{n}+b_{n}}{2}}\right]$ and $I''=\left[{\frac {a_{n}+b_{n}}{2}},b_{n}\right]$ 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 $n\in \mathbb {N}$, it follows by induction that $\left|I_{n}\right|={\frac {\left|I_{1}\right|}{2^{n-1}}}$ for all $n\in \mathbb {N}$. As a consequence, $\left|I_{n}\right|\to 0$ as $n\to \infty$. 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 $k\in \mathbb {N}$. The sequence is built inductively:

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

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

quod erat demonstrandum