Sequences - Calculus Explorer

Definition & Notation

What is a Sequence?

A sequence is an ordered list of numbers that often follow a specific pattern. In calculus, we typically work with infinite sequences, which continue indefinitely.

Mathematically, a sequence can be defined as a function whose domain is the set of natural numbers (or a subset of the natural numbers).

Notation

We denote a sequence using various notations:

\begin{align} \{a_1, a_2, a_3, \ldots, a_n, \ldots\} \\ \{a_n\} \\ \{a_n\}_{n=1}^{\infty} \end{align}

Where:

\(a_1\) is the first term
\(a_2\) is the second term
\(a_n\) is the general term or the \(n^{th}\) term

The general term \(a_n\) is often given by a formula that depends on \(n\).

Example 1: Sequence Notation

For the sequence \(\{a_n\} = \{\frac{n+1}{n^2}\}_{n=1}^{\infty}\), the first few terms are:

\begin{align} a_1 &= \frac{1+1}{1^2} = \frac{2}{1} = 2 \\ a_2 &= \frac{2+1}{2^2} = \frac{3}{4} = 0.75 \\ a_3 &= \frac{3+1}{3^2} = \frac{4}{9} \approx 0.444 \\ a_4 &= \frac{4+1}{4^2} = \frac{5}{16} = 0.3125 \\ a_5 &= \frac{5+1}{5^2} = \frac{6}{25} = 0.24 \end{align}

Graphical Representation

A sequence can be visualized by plotting the points \((n, a_n)\) on a coordinate plane, where \(n\) represents the position in the sequence and \(a_n\) represents the value of the term.

Important Note

When graphing sequences, remember that \(n\) can only take integer values, so the graph consists of discrete points rather than a continuous curve.

Interactive Visualizations

Explore Sequence Behavior

Use the controls below to explore different sequences and observe their behavior as \(n\) increases.

Select a sequence:

Number of terms to display: 20

Sequence Information

Formula: a_n = 1/n

Behavior: Converges to 0

First 5 terms: 1, 0.5, 0.333, 0.25, 0.2

Limits of Sequences

Limit of a Sequence

We say that a sequence \(\{a_n\}\) has a limit \(L\) if the terms of the sequence get arbitrarily close to \(L\) as \(n\) becomes very large.

\(\lim_{n \to \infty} a_n = L\)

Working Definition of Limit

We say that \(\lim_{n \to \infty} a_n = L\) if we can make \(a_n\) as close to \(L\) as we want by taking \(n\) sufficiently large.

Precise Definition of Limit

More formally, \(\lim_{n \to \infty} a_n = L\) if for every \(\varepsilon > 0\), there exists an integer \(N\) such that:

\(|a_n - L| < \varepsilon \text{ whenever } n > N\)

Convergence & Divergence

Convergent Sequence

A sequence \(\{a_n\}\) is said to be convergent if its limit exists and is finite.

\(\lim_{n \to \infty} a_n = L\) where L is a finite number.

Divergent Sequence

A sequence \(\{a_n\}\) is said to be divergent if its limit does not exist or is infinite.

Types of Divergence

A sequence can diverge in several ways:

It can grow without bound: \(\lim_{n \to \infty} a_n = \infty\)
It can decrease without bound: \(\lim_{n \to \infty} a_n = -\infty\)
It can oscillate without approaching any specific value

Properties of Convergent Sequences

If \(\{a_n\}\) and \(\{b_n\}\) are convergent sequences with \(\lim_{n \to \infty} a_n = L\) and \(\lim_{n \to \infty} b_n = M\), then:

\(\lim_{n \to \infty} (a_n + b_n) = L + M\)
\(\lim_{n \to \infty} (a_n - b_n) = L - M\)
\(\lim_{n \to \infty} (a_n \cdot b_n) = L \cdot M\)
\(\lim_{n \to \infty} \frac{a_n}{b_n} = \frac{L}{M}\) (provided \(M \neq 0\) and \(b_n \neq 0\) for all \(n\))

Examples

Example 2: Finding the Limit of a Sequence

Find the limit of the sequence \(\{a_n\} = \{\frac{n+1}{n^2}\}_{n=1}^{\infty}\).

Solution:

\begin{align} \lim_{n \to \infty} \frac{n+1}{n^2} &= \lim_{n \to \infty} \frac{n}{n^2} + \lim_{n \to \infty} \frac{1}{n^2} \\ &= \lim_{n \to \infty} \frac{1}{n} + \lim_{n \to \infty} \frac{1}{n^2} \\ &= 0 + 0 \\ &= 0 \end{align}

Therefore, \(\lim_{n \to \infty} \frac{n+1}{n^2} = 0\).

Example 3: Determining Convergence

Determine whether the following sequences converge or diverge. If they converge, find the limit.

(a) \(\{a_n\} = \{\frac{2n^2 + 3n}{n^2 + 1}\}_{n=1}^{\infty}\)

Solution:

\begin{align} \lim_{n \to \infty} \frac{2n^2 + 3n}{n^2 + 1} &= \lim_{n \to \infty} \frac{2n^2 + 3n}{n^2 + 1} \cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}} \\ &= \lim_{n \to \infty} \frac{2 + \frac{3}{n}}{1 + \frac{1}{n^2}} \\ &= \frac{2 + 0}{1 + 0} \\ &= 2 \end{align}

Note: we have used properties 1, 3, and 4 above.

Therefore, the sequence converges to 2.

(b) \(\{a_n\} = \{(-1)^n\}_{n=1}^{\infty}\)

Solution:

This sequence alternates between -1 and 1:

\(a_1 = -1, a_2 = 1, a_3 = -1, a_4 = 1, \ldots\)

Since the sequence oscillates between two values and does not approach any single value, it diverges.

Solution:

This sequence grows without bound as \(n\) increases:

\(a_1 = 1, a_2 = 2, a_3 = 3, a_4 = 4, \ldots\)

Therefore, \(\lim_{n \to \infty} n = \infty\), and the sequence diverges.

Practice Exercises

Exercise 1

Find the first five terms of the sequence \(\{a_n\} = \{\frac{3n-2}{n+1}\}_{n=1}^{\infty}\).

Exercise 2

Determine whether the following sequences converge or diverge. If they converge, find the limit.

(a) \(\{a_n\} = \{\frac{5n^3 + 2n}{3n^3 + n^2}\}_{n=1}^{\infty}\)

(b) \(\{a_n\} = \{\frac{(-1)^n n}{n+1}\}_{n=1}^{\infty}\)

(a) \(\{a_n\} = \{\frac{5n^3 + 2n}{3n^3 + n^2}\}_{n=1}^{\infty}\)

\begin{align} \lim_{n \to \infty} \frac{5n^3 + 2n}{3n^3 + n^2} &= \lim_{n \to \infty} \frac{5n^3 + 2n}{3n^3 + n^2} \cdot \frac{\frac{1}{n^3}}{\frac{1}{n^3}} \\ &= \lim_{n \to \infty} \frac{5 + \frac{2}{n^2}}{3 + \frac{1}{n}} \\ &= \frac{5 + 0}{3 + 0} \\ &= \frac{5}{3} \end{align}

Therefore, the sequence converges to \(\frac{5}{3}\).

(b) \(\{a_n\} = \{\frac{(-1)^n n}{n+1}\}_{n=1}^{\infty}\)

First, let's examine the behavior of \(\frac{n}{n+1}\):

\(\lim_{n \to \infty} \frac{n}{n+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n}} = 1\)

However, the factor \((-1)^n\) causes the sequence to oscillate between positive and negative values. As \(n\) increases, the terms approach 1 in magnitude but alternate in sign:

\(\lim_{n \to \infty} \left|\frac{(-1)^n n}{n+1}\right| = 1\)

Since the sequence oscillates between values approaching -1 and 1, it diverges.

Let's compute a few terms:

\begin{align} a_1 &= \frac{2^1}{1!} = \frac{2}{1} = 2 \\ a_2 &= \frac{2^2}{2!} = \frac{4}{2} = 2 \\ a_3 &= \frac{2^3}{3!} = \frac{8}{6} = \frac{4}{3} \approx 1.333 \\ a_4 &= \frac{2^4}{4!} = \frac{16}{24} = \frac{2}{3} \approx 0.667 \\ a_5 &= \frac{2^5}{5!} = \frac{32}{120} = \frac{4}{15} \approx 0.267 \end{align}

We can show that \(\lim_{n \to \infty} \frac{2^n}{n!} = 0\) using the Squeeze Theorem. For \(n \geq 4\), the first three factors of the product contribute \(\frac{4}{3}\), and each remaining factor satisfies \(\frac{2}{k} \leq \frac{1}{2}\) for \(k \geq 4\), giving:

\begin{align} 0 \leq \frac{2^n}{n!} &= \underbrace{\frac{2}{1} \cdot \frac{2}{2} \cdot \frac{2}{3}}_{\displaystyle= \,\frac{4}{3}} \cdot \underbrace{\frac{2}{4} \cdots \frac{2}{n}}_{\leq \,\left(\frac{1}{2}\right)^{n-3}} \\[6pt] &\leq \frac{4}{3} \cdot \left(\frac{1}{2}\right)^{n-3} \xrightarrow{n \to \infty} 0 \end{align}

By the Squeeze Theorem, \(\displaystyle\lim_{n \to \infty} \frac{2^n}{n!} = 0\). \(\blacksquare\)

Exercise 3

Prove that if \(\lim_{n \to \infty} a_n = L\) and \(\lim_{n \to \infty} b_n = M\), then \(\lim_{n \to \infty} (a_n + b_n) = L + M\).