3.2 Limit Theorems

<< 3.1 Sequences and Their Limits | 3.3 Monotone Sequences >>

This note covers:

• Boundedness of sequences (3.2.1)
• If a sequence is convergent, then it is bounded (3.2.2)
• Algebra operations on sequences (Definition: Algebra operations on sequences, 3.2.3)
• And various limit theorems:
  • Limit of positive sequence is positive (3.2.4)
  • Order preservation in limit of sequences (3.2.5)
  • Bounds of limits of sequences (3.2.6)
  • Squeeze theorem (3.2.7)
  • Limit of absolute sequences (3.2.9)
  • Limit of square root of sequences (3.2.10)
  • Ratio test if converging to 0 (3.2.11)

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Summary

Definition

• Bounded sequence
• Sequence algebra operations

Theorem

…

3.2.1 Definition: Bounded sequence

A sequence $X=(x_n)$ of real numbers is said to be bounded.

if there exists a real number $M>0$ such that $|x_n|\le M$ for all $n\in \mathbb{N}$.

Thus the sequence of real numbers $(x_n)$ is bounded if and only if the set of its values are a bounded subset of $\mathbb{R}$.

NOTE

To prove a sequence is bounded, find $M$ which $\forall n\in \mathbb{N}$ satisfies $|x_n|<M$.

We can then prove using $-M\le x_n\le M$.

Do example with $|\frac{1}{n}|$!

3.2.2 Theorem: A converging sequence of real numbers is bounded

A converging sequence of real numbers is bounded.

Definition: Algebra operations on sequences

If $X=(x_n)$ and $Y=(y_n)$ a sequence of real numbers, and $c\in \mathbb{R}$ then:

• $X+Y:=(x_n+y_n)$
• $X-Y:=(x_n-y_n)$
• $XY:=(x_n{y}_n)$
• $\frac{X}{Y}:=\left(\frac{x_n}{y_n}\right)$ with $y_n\ne 0,\forall n\in \mathbb{N}$
• $cX:=(c{x}_n)$

3.2.3 Theorem: Convergence and algebra operations

1. Let $X=(x_n)$ and $Y=(y_n)$ be sequences of real numbers that converge to $x$ and $y$, respectively, and let $c\in \mathbb{R}$.

Then the sequences $X+Y,X-Y,X\cdot Y$, and $cX$ converge to $x+y,x-y,xy$, and $cx$, respectively. 2. If $X=(x_n)$ converges to $x$ and $Z=(z_n)$ is a sequence of nonzero real numbers that converges to $z$ and if $\lim (z_n)=z\ne 0$.

Then the quotient sequence $X/Z$ converges to $x/z$.

Notice the difference from the previous definition.

In this theorem, $X$ and $Y$ has to be convergent, so that the resulting limit is also convergent.

We use Theorem 3.1.10 to prove this theorem. See the example attached to that theorem.

3.2.4 Theorem: Limit of positive sequences

If $X=(x_n)$ is a convergent sequence of real numbers and if $x_n\ge 0$ for all $n\in \mathbb{N}$,

then $x=\lim (x_n)\ge 0$.

$X$ is converging, and all its values are positive.

The limit must also be positive.

3.2.5 Theorem: Order preservation in limits of sequences

If $X=(x_n)$ and $Y=(y_n)$ are convergent sequences of real numbers and if $x_n\le y_n$ for all $n\in \mathbb{N}$,

then $\lim (x_n)\le \lim (y_n)$.

In essence similar to 3.2.4.

The difference is instead of comparing with $0$, we’re comparing with another limit.

3.2.6 Theorem: Bounds of limit of sequences

If $X=(x_n)$ is a convergent sequence and if $a\le x_n\le b$ for all $n\in \mathbb{N}$,

then $a\le \lim (x_n)\le b$.

Notion of bounds in converging sequences.

Extension of 3.2.1 and 3.2.2

3.2.7 Theorem: Squeeze theorem

Suppose that $X=(x_n),Y=(y_n)$, and $Z=(z_n)$ are sequences of real numbers such that

$$x_n\le y_n\le z_n,\forall n\in \mathbb{N}$$

and that $\lim (x_n)=\lim (z_n)$.

Then $y=(y_n)$ is convergent and **$$ \lim(x_n) = \lim(y_n) = \lim(z_n)

3.2.9 Theorem: Limit of absolute sequences

Let the sequence $(x_n)$ converge to $x$.

Then the sequence $(|x_n|)$ of absolute values converges to $|x|$. That is, $x=\lim (x_n)\to |x|=\lim (|x_n|)$

From Triangle Inequality.

3.2.10 Theorem: Limit of square root of sequences

Let the sequence $(x_n)$ converge to $x$. and suppose that $x_n\ge 0$.

Then the sequence $(\sqrt{x_n})$ of positive square roots converges to $\sqrt{x}$

3.2.11 Theorem: Ratio test for convergence to zero

Let $(x_n)$ be a sequence of positive real numbers such that $L:=\lim (x_{n+1}/x_n)$ exists.

If $L<1$, then $(x_n)$ converges and $\lim (x_n)=0$.

$L$ is the ratio between two

If all of $(x_n)$ is positive,

and $(x_n)$ is “gradually” decreasing

Proof

3.2.2

1. Premise

Let $(x_n)$ be sequence of real numbers, and $\epsilon :=1$ Suppose that $\lim (x_n)=x$

2. From premise. Using Definition 3.1.3

Then there exists a natural number $K=K(1)$ such that $|x_n-x|<1$ for all $n\ge K$.

3. From 2. Using Triangle Inequality

If we apply the Triangle Inequality with $n\ge K$, we obtain

$$|x_n|=|x_n-x+x|\le |x_n-x|+|x|<1+|x|$$

4. Define.

If we set

$$M:=\sup \{\mid x_1|,|x_2|,\dots ,|x_{K-1}|,1+|x|\},$$

5. From 3, 4.

then it follows that $|x_n|\le M$ for all $n\in \mathbb{N}$.

NOTE

Notice that in $\{\mid x_1|,|x_2|,\dots ,|x_{K-1}|\}$ the subscripts are natural numbers.

Then it follows that the inequality applies for all $n\in \mathbb{N}$

3.2.4

Suppose the conclusion is not true and that $x<0$; then $\epsilon :=-x$ is positive. Since $X$ converges to $x$, there is $K\in \mathbb{N}$ such that

$$x-\epsilon <x_n<x+\epsilon \forall n\ge K$$

In particular, we have $x_K<x+\epsilon =x+(-x)=0$. But this contradicts the hypothesis that $x_n\ge 0$ for all $n\in \mathbb{N}$. Therefore, this contradiction implies that $x\ge 0$.

3.2.5

Let $z_n:=y_n-x_n$ so that $Z:=(z_n)=Y-X$ and $z_n\ge 0$ for all $n\in \mathbb{N}$. It follows from Theorems 3.2.3 and 3.2.4 that

$$0\le \lim Z=\lim (y_n)-\lim (x_n)$$

so that $\lim (x_n)\le \lim (y_n)$

3.2.6

Let $Y$ be the constant sequence $(b,b,b,\dots )$. Theorem 3.2.5 implies that $\lim X\le \lim Y=b$. Similarly one shows that $a\le \lim X$.

3.2.7

3.2.9

It follows from the Triangle Inequality that

$$\left||x_n|-|x|\right|\le |x_n-x|\forall n\in \mathbb{N}$$

Recall the definition of a converging sequence.

The convergence of $(|x_n|)$ to $|x|$ is then an immediate consequence of the convergence of $(x_n)$ to $x$.

See also

• Various Methods for Proving a Sequence of Real Numbers is Convergent