Calculus 2Sequences and Series

Sequences and Their Limits

A sequence converges when its terms approach one number.

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A sequence is an endless list $a_1, a_2, a_3, \ldots$, a function whose inputs are the whole numbers. You met them as recursive formulas. Here the question is where the terms are heading, which is a limit at infinity with $n$ in place of $x$.

Why it works

"$a_n \to L$" means: for any tolerance $\varepsilon > 0$, no matter how small, there is a point in the list after which every term stays within $\varepsilon$ of $L$. If the terms come from a function, $a_n = f(n)$, and $f(x) \to L$ as $x \to \infty$, then $a_n \to L$ too: the sequence just samples the function at whole numbers. That lets you use limit laws and L'Hôpital's rule on sequences.

Key idea

$\lim_{n\to\infty} a_n = L$ means the terms get and stay as close to $L$ as you like. If no such $L$ exists, the sequence diverges. Useful facts: $r^{n} \to 0$ when $|r| < 1$; $\left(1 + \frac{x}{n}\right)^{n} \to e^{x}$; the squeeze theorem works for sequences; if $a_n \to L$ and $g$ is continuous, $g(a_n) \to g(L)$.

Worked example: A rational sequence

Find $\lim_{n\to\infty}\frac{3n^{2} + 1}{2n^{2} - n}$.
  1. Divide top and bottom by $n^{2}$: $\frac{3 + 1/n^{2}}{2 - 1/n}$.
  2. As $n \to \infty$, $\frac{1}{n} \to 0$ and $\frac{1}{n^{2}} \to 0$, so the limit is $\frac{3}{2}$.

Answer: $\frac{3}{2}$

The terms $\frac{3n^{2} + 1}{2n^{2} - n}$ as dots: they start at 4 and settle onto the dashed line at $\frac{3}{2}$.

Worked example: The number e

Find $\lim_{n\to\infty}\left(1 + \frac{2}{n}\right)^{n}$.
  1. Take logs: $n\ln\left(1 + \frac{2}{n}\right) = \frac{\ln(1 + 2/n)}{1/n}$, a $\frac{0}{0}$ form.
  2. L'Hôpital (in $t = \frac{1}{n}$): $\frac{\ln(1 + 2t)}{t} \to 2$ as $t \to 0$.
  3. So the sequence tends to $e^{2}$.

Answer: $e^{2}$

Worked example: Bouncing terms

Does $a_n = (-1)^{n}\frac{n}{n + 1}$ converge?
  1. The size $\frac{n}{n + 1} \to 1$, but the sign flips every step.
  2. So the terms crowd near $1$ for even $n$ and near $-1$ for odd $n$: there is no single number they approach.

Answer: It diverges.

Common mistake

Confusing the limit of the terms with the sum of the terms. For $a_n = \frac{1}{2^{n}}$, the terms $\frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots$ approach $0$; the total $\frac{1}{2} + \frac{1}{4} + \cdots$ approaches 1. Those are two different questions: this topic asks where the terms go; series ask what they add up to.

Practice

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Sources and further reading

  • OpenStax Calculus Volume 2, Section 5.1: Sequences
  • Paul’s Online Math Notes (Lamar University), Calculus II: Sequences
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