3.1 Sequences and Their Limits

<< 2.5 Intervals | 3.2 Limit Theorems >>

This note covers:

• Definition of sequences (3.1.1)
• Limits (convergence, divergence) of sequences (3.1.3)
• Uniqueness of limit of a sequence (3.1.4)
• Tail of a sequence (3.1.8) and its limit (3.1.9)
• One method of finding limit of a sequence (3.1.10)

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Summary

Definition

• Sequences
• Convergent (and divergent) sequences
• Tail of a sequence

Theorem

• $X$: sequence of real numbers
• $X_m$: $m$-tail of $X$

1. $X$ convergent → has exactly one limit
2. $X$ converges to $x$ → $\forall \epsilon >0,\exists K\in \mathbb{N}\ni (|x_n-x|<\epsilon ,\forall n\ge K)$ → $\forall \epsilon >0,\exists K\in \mathbb{N}\ni (x-\epsilon <x_n<x+\epsilon ,\forall n\ge K)$ → $\forall V_{\epsilon (x)},\exists K\in \mathbb{N}\ni (x_n\in V_{\epsilon }(x),\forall n\ge K)$
3. $X_m$ converges ←> $X$ converges
4. See 3.1.10

3.1.1 Definition: Sequence of real numbers

A sequence of real numbers (or sequence in $\mathbb{R}$) is a function with domain $\mathbb{N}$ and codomain $\mathbb{R}$.

In other words, a sequence in $\mathbb{R}$ assigns to each $n\in \mathbb{N}$ a uniquely determined real number. It is usually denoted as: $X,(x_n)(x_n:n\in \mathbb{N})$

The values of $x_n$ are called $n$th-term or elements of the sequence.

A sequence can also be defined by:

1. Formula of $n$-th term:

$$X:=\left(\frac{1}{2n}:n\in \mathbb{N}\right)$$

2. List the terms sequentially, until the pattern seems evident:

$$X:=\left(\frac{1}{2},\frac{1}{4},\frac{1}{6},\frac{1}{8},\dots \right)$$

3. Inductively/recursively

$$x_1:=2,x_{n+1}:=x_n+2$$

4. Constant sequence. e.g., $(0,0,0,\dots )$.

3.1.3 Definition: Converging and diverging (limit) of a sequence

A sequence $(x_n)$ in $\mathbb{R}$ is said to:

• Converge to $x\in \mathbb{R}$ or
• $x$ be a limit of $(x_n)$

if for every $\epsilon >0$ there exists a $K(\epsilon )\in \mathbb{N}$ such that for all $n\ge K(\epsilon )$, the terms $x_n$ satisfy $|x_n-x|<\epsilon$.

If a sequence has a limit, we say it’s convergent. Otherwise, we say it’s divergent.

Drawing-2025-03-10-07.04.00.excalidraw

3.1.8 Definition: Tail of a sequence

If $X=(x_1,x_2,\dots ,x_n,\dots )$ is a sequence of real numbers and if $m$ is a given natural number,

then then the $m$-tail of $X$ is the sequence

$$X_m:=(x_{m+n}:n\in \mathbb{N})=(x_{m+1},x_{m+2},\dots )$$

For example, the 3-tail of the sequence $X=(2,4,6,8,10,\dots ,2n,\dots )$, is the sequence $X_3=(8,10,12,\dots ,2n+6,\dots )$.

3.1.4 Theorem: Uniqueness of limit

A sequence in $\mathbb{R}$ has at most one limit.

$\left[\lim (x_n)=x^{\prime }\wedge \lim (x_n)=x^{\prime \prime }\right]\to x^{\prime }=x^{\prime \prime }$

3.1.5 Theorem: Sequence limit equivalency

Let $(x_n)$ be a sequence of real numbers, and let $x\in \mathbb{R}$. The following statements are equivalent:

1. $X$ converges to $x$.
2. $\forall \epsilon >0$, $\exists K\in \mathbb{N}$ such that $\forall n\ge K$, the terms $x_n$ satisfy $|x_n-x|<\epsilon$.
3. $\forall \epsilon >0$, $\exists K\in \mathbb{N}$ such that $\forall n\ge K$, the terms $x_n$ satisfy $x-\epsilon <x_n<x+\epsilon$.
4. $\forall \epsilon$-neighborhood $V_{\epsilon }(x)$ of $x$, $\exists K\in \mathbb{N}$ such that $\forall n\ge K$, the terms $x_n$ belong to $V_{\epsilon }(x)$.

3.1.9 Theorem: Limit of a tail

Let $(x_n)$ be a sequence of real numbers, and $m\in \mathbb{N}$.

Then the $m$-tail $X_m=(x_{m+n}:n\in \mathbb{N})$ of $X$ converges if and only if $X$ converges.

In this case, $\lim X_m=\lim X$.

3.1.10 Theorem: Finding limit of a sequence

Let $(x_n)$ be a sequence of real numbers and let $x\in \mathbb{R}$. If

• $(a_n)$ is a sequence of positive real numbers
• $\lim (a_n)=0$
• For some constant $C>0$ and $m\in \mathbb{N}$ we have *$$ |x_n - x| \leq Ca_n\qquad\forall n\geq m

then it follows that $\lim (x_n)=x$.

Interpretation

We are basically saying:

For a positive constant $C$, there exists the term $m\in \mathbb{N}$ such that for all terms $n\ge m$, the difference between $x_n$ and $x$ is less than $C{a}_n$.

And because $(a_n)\to 0$, it follows that $|x_n-x|\to 0$, which implies that $(x_n)\to x$

Proofs

3.1.4

1. Premise

Suppose that $x^{\prime }$ and $x^{\prime \prime }$ are both limits of $(x_n)$.

2. From 1, 3.1.3 Definition Converging and diverging (limit) of a sequence

For each $\epsilon >0$,

• there exists $K^{\prime }$ such that $|x_n-x^{\prime }|<\epsilon /2$ for all $n\ge K^{\prime }$ , and
• there exists $K^{\prime \prime }$ such that $|x_n-x^{\prime \prime }|<\epsilon /2$ for all $n\ge K^{\prime \prime }$.

3. Premise

Let $K$ be the larger of $K^{\prime }$ and $K^{\prime \prime }$. i.e., $K=\sup \{K^{\prime },K^{\prime \prime }\}$

4. Obtain $|x^{\prime }-x^{\prime \prime }|=0$ using Triangle Inequality and Theorem: Notion of limit from order.

For $n\ge K$, we apply triangle inequality to get

$$\begin{array}{rl}|x^{\prime }-x^{\prime \prime }|& =|x^{\prime }-x_n+x_n-x^{\prime \prime }|\\ & \le |x^{\prime }-x_n|+|x_n-x^{\prime \prime }|<\epsilon /2+\epsilon /2=\epsilon \end{array}$$

Since $\epsilon >0$ is an arbitrary positive number, we conclude that $|x^{\prime }-x^{\prime \prime }|=0$.

5. Conclusion

Because $|x^{\prime }-x^{\prime \prime }|=0$, we can say that $x^{\prime }=x^{\prime \prime }$.

NOTE

Step 3 is needed because if $K$ is larger than $K^{\prime }$ and $K^{\prime \prime }$,

The 2 statements on step 2 applies for all $n\ge K$ (because $K>K^{\prime }$ and $K>K^{\prime \prime }$.

3.1.5

Points 1. and 2. are just the definition.

Point 3 is obtained using Theorem 2.2.2 (3).

$$\begin{array}{rl}|& x_n-x|<\epsilon \\ -\epsilon <& x_n-x<\epsilon \\ x-\epsilon <& x_n-x<x+\epsilon \end{array}$$

Point 4 is obtained using Definition 2.2.7: Neighborhood. We are basically saying $x_n$ is “close” to $x$.

3.1.9

1. From 3.1.8 Definition Tail of a sequence

Note that for any $p\in \mathbb{N}$, the $p$-th term of $X_m$ is the $(p+m)$-th term of $X$. Similarly, if $q>m$, then the $q$-th term of $X$ is the $(q-m)$-th term of $X_m$.

2. Premise

Assume $X$ converges to $X$.

3. 3.1.3 Definition Converging and diverging (limit) of a sequence

Then given any $\epsilon >0$, if the terms of $X$ for $n\ge K(\epsilon )$ satisfy $|x_n-x|<\epsilon$,

4. From 1.

then the terms of $X_m$ for $k\ge K(\epsilon )-m$ satisfy $|x_k-x|<\epsilon$. Thus we can take $K_m(\epsilon )=K(\epsilon )-m$, so that $X_m$ also converges to $x$.

5. Same as previous 2 steps

Conversely, if the terms of $X_m$ for $k\ge K_m(\epsilon )$ satisfy $|x_k-x|<\epsilon$, then the terms of $X$ for $n\ge K(\epsilon )+m$ satisfy $|x_n-x|<\epsilon$. Thus we can take $K(\epsilon )=K_m(\epsilon )+m$.

6. Conclusion

Therefore, $X$ converges to $x$ if and only if $X_m$ converges to $x$.

3.1.10

1. Premise

• $(a_n)$ sequence of positive real numbers
• $\lim (a_n)=0$
• $C>0$, $m\in \mathbb{N}$
• $|x_n-x|\le C{a}_n,\forall n\ge m$

2. From 1,

a_n &= |a_n| && \text{Premise 1}\\ &= |a_n - 0| && (\text{Addition identity})\\ &< \varepsilon && (\text{Definition of limit of a sequence})\\ &< \varepsilon / C && (C > 0, \varepsilon > 0) \end{align}$$

3. From premise,

|x_n - x| &\leq Ca_n && \text{Premise 4}\\ &<C \left( \frac \varepsilon C \right) && a_n < \varepsilon / C\\ &<\varepsilon \end{align}$$

4. Conclusion using 3.1.3 Definition: Converging and diverging (limit) of a sequence

Since $\epsilon >0$ is arbitrary, we conclude that $x_n=\lim (x_n)$

Example

See Examples