Sequences (monotonic, bounded and convergent)

Definition of a sequence

An infinite list of numbers.

What function represents a sequence of real numbers?

f:\mathbb{N}\to\mathbb{R}

what function represents a sequence of complex numbers?

f:\mathbb{N}\to\mathbb{C}

How can we represent a sequence of complex numebrs on a graph?

Plot the image of the function in an Argand diagram or plot the graph of the function by displaying an Argand diagram for each natural number.

What does it mean for a sequence to be increasing?

a_n is increasing if and only if for all n\in\mathbb{N} we have a_n+1 \ge a_n

The same for decreasing but reversed

What is a monotonic sequence?

A sequence is monotonic if and only if a_n is either increasing or decreasing

Definition of bounded above

A sequence is bounded above if and only if there exists M\in\mathbb{R} such that for all n\in\mathbb{N} we have a_{n}\le M

Definition of bounded below

A sequence is bounded below if and only if there exists M\in\mathbb{R} such that for all n\in\mathbb{N} we have M\le a_{n}

Definition of bounded

if and only iff there exists M\in\mathbb{R} such that for all n\in\mathbb{N} we have |a_{n}|\le M

Unbounded definition

if and only if for all M\in\mathbb{R} there exists n\in\mathbb{N} such that |a_{n}|>M

Definition of bounded for a sequence of complex numbers

if and only if there exists M\in\mathbb{R} such that for all n\in\mathbb{N} we have |z_{n}|\le M

Definition of a convergent sequence

We say a_n → a as n→\infty if and only if for all \varepsilon>0 there exists N\in\mathbb{N} such that for all n\in\mathbb{N} if n\ge N then |a_{n}-a|<\varepsilon.

definition of convergence for complex numbers

A sequence of complex numbers converges to a if and only if for all \varepsilon>0 there exists N\in\mathbb{N} such that for all n\in\mathbb{N} if n\ge N then |a_{n}-a|<\varepsilon.

Whats an easier way to work out if a sequence of complex numbers converges?

Let (a_n) be a sequence of complex numbers. Let (x_n)=(Re(a_n) and let (y_n)=(Im(a_n ) be the sequences of the real and imaginary parts of a_n respectively. Then a_n converges if and only if

• Suppose a_n→a as n→\infty, then x_n → Re(a)and y_n→Im(a) as n→\infty

• a_n → x+iy

What is the uniqueness of limits say for real sequences?

If (a_n) is a real sequence and (a_n)\to L and (a_n)\to M as n'\to\infty then L=M

• In other words, a convergent sequence has at most one limit

How do you prove uniqueness of limits for a sequence?

Suppose that a_n \to a and a_n \to a’ as n \to \infty. We show that a = a’.

To see this, suppose for a contradiction that a \neq a’.

Let \varepsilon = |a - a’|/3. Then \varepsilon > 0.

As a_n \to a as n \to \infty, there exists N_1 \in \mathbb{N} such that if n \ge N_1 then

|a_n - a| < \varepsilon.

Similarly, as a_n \to a’ as n \to \infty, there exists N_2 \in \mathbb{N} such that if n \ge N_2 then

|a_n - a’| < \varepsilon.

Let N = \max{N_1, N_2}. Then

|a - a'| = |(a - a_N) + (a_N - a')| \le |a - a_N| + |a_N - a'| < \tfrac13|a - a'| + \tfrac13|a - a'| = \tfrac23|a - a'|

This shows that

|a - a'| < \tfrac23 |a - a'|

which is impossible.

Hence our assumption that a \neq a’ must be false.

Therefore a = a’.

Is every convergent sequence bounded?

Yes.

Suppose (a_n)\to a, as n\to\infty.

Take \varepsilon=1 in the definition of convergence, so \exists N\in\mathbb N such that n\ge N \Rightarrow |a_n-a|<1.

Then for n\ge N, |a_n|=|(a_n-a)+a|\le |a_n-a|+|a|<1+|a|, so the tail is bounded. Let M=\max\{|a_1|,\dots,|a_{N-1}|,\,1+|a|\}. Then |a_n|\le M, for all n\in\mathbb N, hence (a_n) is bounded.

What can we say about An if it is increasing and bounded above?

Then it converges

Let (a_n) be increasing and bounded above, so \exists M\in\mathbb R with a_n\le M for all n.

Set A=\left\lbrace{a_n\mid n\in\mathbb N}\rbrace\right. , then A is non-empty and bounded above by M, so by the completeness axioms of real numbers \sup A=a exists.

First, a_n\le a for all n since a is an upper bound.

Let \varepsilon>0.

Claim: \exists N\in\mathbb N with a-\varepsilon<a_N\le a; otherwise a_N<a-\varepsilon for all N, so a-\varepsilon is an upper bound of A, contradicting least-upper-boundness of a.

With such N, monotonicity gives a_N\le a_n for all n\ge N, hence a-\varepsilon<a_n\le a+\varepsilon for all n\ge N, so |a_n-a|<\varepsilon. Therefore a_n\to a and (a_n) converges

Definition of a subsequence?

A subsequence of a_{n}, is a new sequence a_{n_{1}},a_{n_{2}},a_{n_{3}}… formed by taking some terms of a_{n} in the same order

Denoted by (a_{n})_{j},j\in\mathbb{N}

Proof that a convergent sequence and all its subsequences have the same limit

Assume a_n\to a, as n\to\infty, and let (a_{n_j})_{j\in\mathbb N} be a subsequence with n_1<n_2<\cdots.

Let \varepsilon>0. Since a_n\to a, \exists N\in\mathbb N such that n\ge N\Rightarrow |a_n-a|<\varepsilon.

Choose J\in\mathbb N with n_J\ge N; then for all j\ge J we have n_j\ge n_J\ge N, hence |a_{n_j}-a|<\varepsilon.

Therefore a_{n_j}\to a, as j\to\infty

Whats the limit point of a sequence?

A point to which some subsequence of a_{n} converges.

How can we prove that a sequence doesn’t converge using limit points?

If a sequence has more than one limit then the sequence doesn’t converge.

What is a peak term of a sequence?

a_{r} is called a peak term iff a_{r}>a_{s} for all s>r .

Prove every real sequence has a monotone subsequence.

Let (a_n) be a real sequence. If there are infinitely many peak terms a_{n_1},a_{n_2},\dots, then a_{n_1}>a_{n_2}>\cdots, so we have a decreasing subsequence. If there are only finitely many peak terms, pick n_1 after the last peak term (or any n_1 if none). Since a_{n_1} is not a peak, \exists n_2>n_1 with a_{n_2}\ge a_{n_1}. Again a_{n_2} is not a peak, so \exists n_3>n_2 with a_{n_3}\ge a_{n_2}. Continuing inductively gives an increasing subsequence (a_{n_j}). Hence every real sequence has a monotone subsequence.

Proof of if a_{n} is bounded then there exists a convergent subsequence.

Let (a_n) be a bounded real sequence. Then (a_n) has a monotone subsequence (a_{n_k}).

Since (a_n) is bounded, every subsequence is also bounded, so (a_{n_k}) is bounded and monotone.

Every bounded monotone sequence is convergent. Hence (a_{n_k}) converges, so (a_n) has a convergent subsequence. □

What can we say about a monotone sequence that is bounded.

It is convergent

Definition of a divergent sequence

a_{n} diverges to +\infty as n\to\infty if and only iff for all k>0 there exists N\in\mathbb{N} such that for all n\in\mathbb{N} if n\ge N then a_{n}\ge K.

Proof that if a_{n} is increasing and not bounded above then a_{n} diverges to +\infty

Let k>0 be given.

As a_{n} isn’t bounded above, \exists N\in\mathbb{N} S.T a_{N}>k. As a_{n} is increasing, for all n\ge N we have a_{n} \ge a_{N} \ge k. Hence a_{n} diverges to +\infty

prove if (a_n) is a sequence and a_n\to a and c is a constant, then the sequence (c a_n) converges and c a_n\to c a.

Let \varepsilon>0.

If c=0 then |c a_n-c a|=0<\varepsilon for all n.

If c\neq 0, since a_n\to a there exists N\in\mathbb{N} such that for all n\ge N, |a_n-a|<\varepsilon/|c|. Then for all n\ge N, |c a_n-c a|=|c||a_n-a|<\varepsilon, hence c a_n\to c a."

If (a_n)\to a and (b_n)\to b, what can be said about the sequence (a_n+b_n)?

The sequence (a_n+b_n) converges and a_n+b_n\to a+b.

Proof: Let \varepsilon>0. Since a_n\to a there exists N_1\in\mathbb{N} such that for all n\ge N_1, |a_n-a|<\varepsilon/2, and since b_n\to b there exists N_2\in\mathbb{N} such that for all n\ge N_2, |b_n-b|<\varepsilon/2.

Let N=\max\{N_1,N_2\}. Then for all n\ge N, |(a_n+b_n)-(a+b)|\le |a_n-a|+|b_n-b|<\varepsilon, hence a_n+b_n\to a+b."

If (a_n)\to a and (b_n)\to b, what can be said about the sequence (a_nb_n)and proof?

The sequence (a_nb_n) is convergent and a_nb_n\to ab.

Proof: Let \varepsilon>0. Since (b_n) is convergent, it is bounded, so \exists K>0 such that |b_n|\le K for all n\in\mathbb{N}.

Choose N_1\in\mathbb{N} such that n\ge N_1\Rightarrow |a_n-a|<\varepsilon/(2K).

Choose N_2\in\mathbb{N} such that n\ge N_2\Rightarrow |b_n-b|<\varepsilon/(2(|a|+1)).

Let N=\max\{N_1,N_2\}.

For n\ge N, |a_{n}b_{n}-ab|=|(a_{n}b_{n}-ab_{n})+(ab_{n}-ab)|

\le|a_{n}-a||b_{n}|+|a||b_{n}-b|

<\varepsilon k/2k+\frac{|a|\varepsilon}{2(|a|+1))}<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon

a_nb_n\to ab."

If (b_n) is convergent with b_n\to b , and b_n\neq 0 for all n\ge 1 and b\neq 0, what can be said about the sequence c_n=\frac{1}{b_n}?

(c_n) is convergent and c_n\to \frac{1}{b}.

Proof: Let \varepsilon>0.

If b_n\to b and b_n\neq 0 then there exist K>0 and N_1\in\mathbb{N} such that for all n\ge N_1 we have 0<K<|b_n|

Hence \frac{1}{|b_n|}<\frac{1}{K}.

Since b_n\to b, there exists N_2\in\mathbb{N} such that for all n\ge N_2, |b_n-b|<K|b|\varepsilon

Let N=\max\{N_1,N_2\}. Then for all n\ge N, \left|\frac{1}{b_n}-\frac{1}{b}\right|=\frac{|b_n-b|}{|b_n||b|}<\frac{1}{K|b|}|b_n-b|<\frac{1}{K|b|}\cdot K|b|\varepsilon=\varepsilon, so \frac{1}{b_n}\to \frac{1}{b}.

If (a_n)\to a and (b_n)\to b with b\neq 0 and b_n\neq 0 for all sufficiently large n, what can be said about the sequence \left(\frac{a_n}{b_n}\right)?

The sequence \left(\frac{a_n}{b_n}\right) is convergent and \frac{a_n}{b_n}\to \frac{a}{b}.

Proof:

Since b_n\to b\neq 0, by previous results we have \frac{1}{b_n}\to \frac{1}{b}. By previous results the product of convergent sequences is convergent, so a_n\cdot\frac{1}{b_n}\to a\cdot\frac{1}{b}. Hence \frac{a_n}{b_n}\to \frac{a}{b}."

If (a_n),(b_n),(c_n) are sequences of real numbers with a_n\le b_n\le c_n for all n\in\mathbb{N} and a_n\to a and c_n\to a, what must b_n converge to?

b_n\to a as n\to\infty.

Proof: Let \varepsilon>0.

Since a_n\to a, there exists N_1\in\mathbb{N} such that if n\ge N_1 then |a_n-a|<\varepsilon, i.e. a-\varepsilon<a_n<a+\varepsilon.

Since c_n\to a, there exists N_2\in\mathbb{N} such that if n\ge N_2 then |c_n-a|<\varepsilon, i.e. a-\varepsilon<c_n<a+\varepsilon. Let N=\max\{N_1,N_2\}.

For n\ge N we have a-\varepsilon<a_n\le b_n\le c_n<a+\varepsilon, so |b_n-a|<\varepsilon. Hence b_n\to a."

What does the ratio test say for a sequence (a_n) .

If \frac{a_{n+1}}{a_n}\to \ell as n\to\infty and |\ell|<1, then a_n\to 0.

What does it mean for a sequence (a_n) (of real or complex numbers) to be Cauchy?

(a_n) is a Cauchy sequence if and only if for all \varepsilon>0 there exists N\in\mathbb{N} such that for all n,m\in\mathbb{N}, if n,m\ge N then |a_n-a_m|<\varepsilon.

If a sequence (a_n) converges, what can you say about it in terms of the Cauchy property?

Every convergent sequence is Cauchy.

Proof: Suppose a_n\to a. Let \varepsilon>0. Choose N\in\mathbb{N} such that for all n\ge N, |a_n-a|<\varepsilon/2. Then for all n,m\ge N, |a_n-a_m|\le |a_n-a|+|a_m-a|<\varepsilon/2+\varepsilon/2=\varepsilon, so (a_n) is Cauchy."

What does it mean for a subset A\subseteq \mathbb{R} or A\subseteq \mathbb{C} to be complete?

A is complete if and only if every Cauchy sequence with terms in A converges to a limit in A.

proof that if a sequence is cauchy then it is convergent.

Let (a_n) be a Cauchy sequence.

A Cauchy sequence is bounded. Take \varepsilon=1. Then there exists N such that for all n,m\ge N, |a_n-a_m|<1. Fix m=N.

For all n\ge N,|a_n|=|(a_n-a_N)+a_N|\le |a_n-a_N|+|a_N|<1+|a_N|

Hence the sequence is bounded.

Step 2:

A bounded sequence has a convergent subsequence so there exists a subsequence of (a_n) that converges to some a\in\mathbb R.

Step 3:

The whole sequence converges to the same limit. Let \varepsilon>0. Since the subsequence converges to a, there exists J\in\mathbb{N} such that if j\ge J then |a_{n_{j}}-a|<\varepsilon/2

Also a_{n} is cauchy so there exists N_1 such that for all n,m\ge N_1 , |a_n-a_m|<\varepsilon/2. Let N=\max\{N_1,n_{J}\}.

Suppose that n\ge N , choose J\in\mathbb{N} such that n_{J}\ge N.

Then we have,

|a_{n}-a|\le|a_{n}-a_{n_{J}}|+|a_{n_{J}}-a|<\varepsilon therefore a_n\to a, so the sequence is convergent.