Math Core

Before you can add up infinitely many numbers, you need a precise way to talk about an infinite list of numbers and where it is heading. That list is a sequence, and the question "where is it heading?" is a limit question you already know how to answer.

What a sequence is

A sequence is an ordered, never-ending list of numbers:

a_1, \; a_2, \; a_3, \; \dots, \; a_n, \; \dots

The number $a_n$ is the $n$ th term. Most sequences you meet on the AP exam are given by an explicit formula. For example, $a_n = \dfrac{n}{n+1}$ produces

\frac{1}{2}, \; \frac{2}{3}, \; \frac{3}{4}, \; \frac{4}{5}, \; \dots

You can think of a sequence as a function whose inputs are only the positive integers, $a_n = f(n)$ . Its graph is a string of separate dots rather than a curve. Sometimes the index starts at $n = 0$ instead of $n = 1$ ; the starting point never affects whether the sequence converges.

A sequence can also be defined recursively, by giving a first term and a rule for getting each term from the one before, such as $a_1 = 1$ and $a_{n+1} = 2a_n + 1$ , which gives $1, 3, 7, 15, \dots$

Convergence

As $n$ grows, the terms of $\dfrac{n}{n+1}$ creep closer and closer to $1$ . That is exactly what a limit at infinity describes.

Definition

Convergent sequence

A sequence $\{a_n\}$ converges to the number $L$ if $a_n$ gets arbitrarily close to $L$ for all sufficiently large $n$ . We write

\lim_{n \to \infty} a_n = L.

If no such finite number $L$ exists, the sequence diverges.

A sequence can diverge in two different ways. It can grow without bound, like $a_n = n^2$ (we say it diverges to $\infty$ ), or it can bounce around forever without settling, like $a_n = (-1)^n$ , which alternates $-1, 1, -1, 1, \dots$

The terms of a_n = n/(n+1) climb toward the horizontal line y = 1.Open in grapher →

Tools for finding the limit

The big advantage of the function point of view is this: if $a_n = f(n)$ and $\displaystyle\lim_{x \to \infty} f(x) = L$ , then $\displaystyle\lim_{n \to \infty} a_n = L$ as well. So every limit technique you already have still works.

Finding the limit of a sequence

Rational expressions: compare the highest powers of $n$ in the numerator and denominator.
Indeterminate forms such as $\dfrac{\infty}{\infty}$ : rewrite $a_n$ as $f(x)$ and use L'Hospital's Rule.
Growth rates: for large $n$ , $\ln n \ll n^p \ll b^n \ll n!$ (for any $p > 0$ and $b > 1$ ). A slower-growing quantity divided by a faster one goes to $0$ .
Squeeze Theorem: if $b_n \le a_n \le c_n$ and both $b_n$ and $c_n$ approach $L$ , so does $a_n$ .
Absolute value: if $|a_n| \to 0$ , then $a_n \to 0$ .

A few limits come up so often that they are worth memorizing:

\lim_{n\to\infty} \frac{1}{n^p} = 0 \;(p > 0), \qquad \lim_{n\to\infty} r^n = 0 \;(|r| < 1), \qquad \lim_{n\to\infty} \left(1 + \frac{k}{n}\right)^n = e^k.

Worked examples

Worked example: A rational sequence

Determine whether $a_n = \dfrac{5n^2 - 3}{2n^2 + 7n}$ converges, and if so, find its limit.

Solution. The highest power in both the numerator and the denominator is $n^2$ . Divide every term by $n^2$ :

a_n = \frac{5 - \frac{3}{n^2}}{2 + \frac{7}{n}} \;\longrightarrow\; \frac{5 - 0}{2 + 0} = \frac{5}{2}.

The sequence converges to $\dfrac{5}{2}$ .

Worked example: Using L'Hospital's Rule

Find $\displaystyle\lim_{n\to\infty} \frac{n^2}{e^n}$ .

Solution. This has the form $\dfrac{\infty}{\infty}$ . Let $f(x) = \dfrac{x^2}{e^x}$ and apply L'Hospital's Rule twice:

\lim_{x\to\infty} \frac{x^2}{e^x} = \lim_{x\to\infty} \frac{2x}{e^x} = \lim_{x\to\infty} \frac{2}{e^x} = 0.

So the sequence converges to $0$ , which matches the growth-rate rule: exponentials beat powers.

Worked example: An oscillating sequence

Does $a_n = \dfrac{(-1)^n \, n}{n + 4}$ converge?

Solution. The size of the terms, $\dfrac{n}{n+4}$ , approaches $1$ . So for large $n$ the even terms are close to $1$ and the odd terms are close to $-1$ . The terms never settle on a single value, so the sequence diverges.

Compare this with $b_n = \dfrac{(-1)^n}{n}$ . Here $|b_n| = \dfrac{1}{n} \to 0$ , so $b_n \to 0$ and the sequence converges.

Worked example: A recursive sequence

The sequence defined by $a_1 = 1$ and $a_{n+1} = \dfrac{1}{2}a_n + 4$ is known to converge. Find its limit.

Solution. If $a_n \to L$ , then $a_{n+1} \to L$ too. Take the limit of both sides of the recursion:

L = \frac{1}{2}L + 4 \quad\Longrightarrow\quad \frac{1}{2}L = 4 \quad\Longrightarrow\quad L = 8.

(The first few terms are $1, 4.5, 6.25, 7.125, \dots$ , which are indeed approaching $8$ .)

Common mistake

Do not confuse a sequence with a series. The sequence $\dfrac{1}{n}$ converges to $0$ . The series $1 + \dfrac{1}{2} + \dfrac{1}{3} + \cdots$ , which adds those terms, turns out to diverge. A sequence converging to $0$ does not mean the sum of its terms is finite. You will see why in the next lessons.

Tip

When you suspect a sequence oscillates, look at $|a_n|$ first. If $|a_n| \to 0$ , the sequence converges to $0$ . If $|a_n|$ approaches a positive number while the sign keeps flipping, the sequence diverges.