5.1 Sequences

Big idea

A sequence is a list of numbers with a first entry, a second entry, and so on forever. Formally it is a function whose domain is the positive integers, and that description is the useful one: everything you know about limits of functions transfers, with one adjustment for the fact that the input now jumps rather than slides.

The only question that matters in this section is whether the list settles down. A sequence converges when its terms eventually crowd around a single number and stay there, and the number is its limit. A sequence that fails to do this diverges, whether it runs off to infinity, oscillates between two values, or wanders without pattern. Notice that convergence is a statement about the tail: changing the first thousand terms changes nothing.

This is the foundation of the whole chapter, because an infinite series will be defined as the limit of a sequence, namely its sequence of partial sums. Every convergence test you meet later is ultimately a statement about a sequence, so the tools here are not preliminaries to be skimmed. They are the tools.

There are three of them worth keeping. When a sequence comes from a formula that also makes sense for real inputs, the limit of the function gives the limit of the sequence, which puts L’Hopital’s rule at your disposal. When the terms are squeezed between two sequences with a common limit, they inherit it. And when a sequence is increasing and can never exceed some ceiling, it must converge, even when you have no formula for the limit.

Decoder

A sequence converges to a number when every open interval around that number, however small, contains all terms of the sequence from some index onward.

The phrase “from some index onward” is where the content lives. It does not ask the terms to enter the interval and never leave in the ordinary sense of a first visit; it asks that there be a point in the list after which no term is ever outside again. That is why finitely many terms can be anywhere at all without affecting the answer.

Because the domain is the integers, a sequence has no chance to oscillate faster and faster the way a function can. But it can still oscillate: $a_n = (-1)^n$ has terms that never settle, so it diverges even though both values it visits are perfectly ordinary numbers. Divergence does not require growth.

The standard mistake is to conclude convergence from terms getting smaller. Terms shrinking toward zero says the sequence converges to $0$, and that is fine, but the same intuition applied to a series is false and expensive. Keep the two objects separate: in this section the terms are the sequence.

Definitions and results

Sequence. An ordered list $a_1, a_2, a_3, \dots$, written $\{a_n\}$. It may be given by an explicit formula for $a_n$ or by a recursion that builds each term from earlier ones.

Convergence and divergence. $\{a_n\}$ converges to $L$, written $\lim_{n \to \infty} a_n = L$, when the terms are eventually within any prescribed distance of $L$. If no such $L$ exists, the sequence diverges.

Limit laws. If $\{a_n\}$ and $\{b_n\}$ converge, then sums, differences, products, and constant multiples converge to the corresponding combination of limits, and quotients do too provided the limit of the denominator is not $0$.

Limits from functions. If $f$ is defined for real $x \ge 1$, if $\lim_{x \to \infty} f(x) = L$, and if $a_n = f(n)$, then $\lim_{n \to \infty} a_n = L$. This licenses L’Hopital’s rule on sequences, applied to the function first. The converse fails: the function may oscillate between integers without the sequence noticing.

Squeeze theorem. If $b_n \le a_n \le c_n$ for all large $n$ and $\{b_n\}$ and $\{c_n\}$ share the limit $L$, then $\{a_n\}$ converges to $L$.

Absolute value test. If $\lim_{n \to \infty} |a_n| = 0$ then $\lim_{n \to \infty} a_n = 0$. This is the squeeze theorem applied to $-|a_n| \le a_n \le |a_n|$, and it is the fastest way to handle terms with alternating signs.

Geometric sequences. For $\{r^n\}$: the limit is $0$ when $|r| < 1$, the limit is $1$ when $r = 1$, and the sequence diverges for every other $r$, including $r = -1$.

Monotone and bounded. A sequence is increasing if $a_{n+1} \ge a_n$ for all $n$ and decreasing if $a_{n+1} \le a_n$; either way it is monotone. It is bounded above if some number exceeds every term, bounded below if some number is under every term. The monotone convergence theorem says a monotone bounded sequence converges. The theorem produces existence, not a value.

Growth comparisons. For large $n$, factorials beat exponentials and exponentials beat powers. In particular $\frac{r^n}{n!} \to 0$ for every fixed $r$, and $\frac{\ln n}{n^p} \to 0$ for every $p > 0$.

Worked examples

A ratio of polynomials in the index

Find $\lim_{n \to \infty} \dfrac{3n^2 - n}{2n^2 + 5}$.

Divide numerator and denominator by the highest power present, which is $n^2$:

$$ \frac{3n^2 - n}{2n^2 + 5} = \frac{3 - \frac{1}{n}}{2 + \frac{5}{n^2}} $$

Both $\frac1n$ and $\frac{5}{n^2}$ go to $0$, so by the limit laws the value is $\frac{3}{2}$. The sequence converges. As a sanity check, $n = 100$ gives $\frac{29900}{20005} \approx 1.4946$, already close to $1.5$.

A limit that needs L’Hopital’s rule

Find $\lim_{n \to \infty} \dfrac{\ln n}{\sqrt{n}}$.

The sequence itself cannot be differentiated, so pass to the function $f(x) = \frac{\ln x}{\sqrt x}$, which agrees with the sequence at the integers. Both parts go to infinity, so the rule applies:

$$ \lim_{x \to \infty}\frac{\ln x}{\sqrt x} = \lim_{x \to \infty}\frac{1/x}{1/(2\sqrt x)} = \lim_{x \to \infty}\frac{2\sqrt{x}}{x} = \lim_{x \to \infty}\frac{2}{\sqrt x} = 0 $$

Since the function tends to $0$, so does the sequence. This is the general pattern behind “logarithms lose to powers”.

Oscillating terms, two different outcomes

Consider $a_n = \dfrac{\sin(n^2)}{n}$ and $b_n = (-1)^n\dfrac{n}{n+1}$.

For the first, there is no hope of evaluating $\sin(n^2)$, but there is no need. Since $|\sin(n^2)| \le 1$,

$$ 0 \le \left|\frac{\sin(n^2)}{n}\right| \le \frac{1}{n} $$

and the right side goes to $0$, so $|a_n| \to 0$ and therefore $a_n \to 0$.

For the second, the sizes $\frac{n}{n+1}$ approach $1$ while the sign flips every step, so the terms crowd near $1$ and near $-1$ and never near a single value. No interval of radius $\frac12$ around any candidate can hold the whole tail, so $\{b_n\}$ diverges. The difference between the two is that the first has factors shrinking to zero and the second does not.

A recursion handled without a formula

Let $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$. The first few terms are $1$, $\sqrt3 \approx 1.7321$, $\sqrt{3.7321} \approx 1.9319$, $\sqrt{3.9319} \approx 1.9829$.

First show every term is below $2$. It holds for $a_1 = 1$, and if $a_n < 2$ then $a_{n+1} = \sqrt{2 + a_n} < \sqrt{4} = 2$. So the sequence is bounded above by $2$.

Next show it increases. Since all terms are positive, $a_{n+1} > a_n$ is equivalent to $2 + a_n > a_n^2$, that is $a_n^2 - a_n - 2 < 0$, that is $(a_n - 2)(a_n + 1) < 0$. Because $0 < a_n < 2$, the first factor is negative and the second positive, so the product is negative and the inequality holds.

Monotone and bounded, so the sequence converges; call the limit $L$. Taking limits on both sides of the recursion, and using that $\{a_{n+1}\}$ has the same limit as $\{a_n\}$, gives $L = \sqrt{2 + L}$. Squaring, $L^2 - L - 2 = 0$, so $(L - 2)(L + 1) = 0$ and $L = 2$ or $L = -1$. The terms are positive, so $L = 2$.

Note the order of the argument. Convergence was established first and the equation was solved second. Solving $L = \sqrt{2 + L}$ on its own proves nothing, since a divergent sequence would satisfy no such equation to begin with.

Practice

One skill, covering the full range: evaluating limits of explicit sequences, deciding convergence or divergence, applying the squeeze theorem to oscillating terms, and using monotonicity with a bound on recursive sequences.

Quiz

Five items on computing sequence limits, recognizing divergence, and applying the squeeze and monotone convergence theorems.