Convergence of Sequences, calculus lesson

Press the square root key on a calculator again and again, starting from any positive number other than $1$ , and the display creeps toward $1$ . It never gets there, but after enough presses it is as close to $1$ as the display can show. Convergence makes that precise. However small a tolerance you name, the terms eventually stay within it for good.

Key terms

The epsilon N definition

Definition A sequence $\{a_n\}$ converges to the real number $L$ , written $\lim_{n \to \infty} a_n = L$ or $a_n \to L$ , if for every $\varepsilon > 0$ there is an integer $N$ such that

\lvert a_n - L \rvert < \varepsilon \quad \text{for all } n > N.

In plain words Pick any tolerance $\varepsilon$ , as small as you like. From some point on, every term is within $\varepsilon$ of $L$ . A smaller tolerance may need a later starting point $N$ , and that is allowed.

Example For $a_n = (n + 1)/n$ and $L = 1$ we have

\lvert a_n - 1 \rvert = \left\lvert \frac{n + 1}{n} - 1 \right\rvert = \frac{1}{n}.

This is less than $\varepsilon$ when $n > 1/\varepsilon$ , so any integer $N \ge 1/\varepsilon$ works. For $\varepsilon = 0.01$ take $N = 100$ . Indeed $a_{101} = \frac{102}{101} \approx 1.0099$ is within $0.01$ of $1$ .

Finite limits

Definition A sequence is convergent if it converges to some real number $L$ , called its limit. The limit of a convergent sequence is unique. If $\lim_{x \to \infty} f(x) = L$ and $a_n = f(n)$ for all integers $n \ge 1$ , then $\lim_{n \to \infty} a_n = L$ . If $a_n \to L$ , $b_n \to M$ and $c$ is a constant, then

a_n + b_n \to L + M, \qquad c\,a_n \to cL, \qquad a_n b_n \to LM,

and if moreover $M \neq 0$ and $b_n \neq 0$ for all $n$ , then

\frac{a_n}{b_n} \to \frac{L}{M}.

In plain words A convergent sequence settles on exactly one number, and you can find that number with the same algebra you used for limits of functions at infinity.

Example Dividing top and bottom by $n^2$ ,

\lim_{n \to \infty} \frac{3n^2 + n}{n^2 + 4} = \lim_{n \to \infty} \frac{3 + \frac{1}{n}}{1 + \frac{4}{n^2}} = \frac{3 + 0}{1 + 0} = 3.

Divergence

Definition A sequence diverges if it does not converge to any real number. It diverges to $\infty$ , written $a_n \to \infty$ , if for every number $M$ there is an integer $N$ with $a_n > M$ for all $n > N$ . Divergence to $-\infty$ is defined the same way with $a_n < M$ .

In plain words Divergent just means not convergent. Running off to infinity is one way to diverge, and bouncing around forever is another.

Example For $a_n = \sqrt{n}$ and any $M > 0$ , if $n > M^2$ then $\sqrt{n} > M$ . So $N = \lceil M^2 \rceil$ works and $\sqrt{n} \to \infty$ . For $M = 100$ every term after $a_{10000}$ exceeds $100$ .

Unboundedness

Definition Every convergent sequence is bounded. Here is why. If $a_n \to L$ , take $\varepsilon = 1$ to get $N$ with $\lvert a_n \rvert < \lvert L \rvert + 1$ for $n > N$ . Then

\lvert a_n \rvert \le \max\{\lvert a_1 \rvert, \dots, \lvert a_N \rvert, \lvert L \rvert + 1\} \quad \text{for all } n.

Consequently an unbounded sequence diverges. The converse is false, since a bounded sequence need not converge.

In plain words If the terms are not trapped in any band, they cannot be settling on one number.

Example The sequence $a_n = (-2)^n$ is $-2, 4, -8, 16, \dots$ , and $\lvert a_n \rvert = 2^n$ exceeds every bound. So it diverges. It does not diverge to $\infty$ or to $-\infty$ , since its terms keep changing sign.

Oscillation

Definition A bounded sequence can still diverge if its terms keep coming back near two or more different values. A subsequence of $\{a_n\}$ is a sequence $a_{n_1}, a_{n_2}, a_{n_3}, \dots$ whose indices are strictly increasing, $n_1 < n_2 < n_3 < \cdots$ . If $a_n \to L$ , then every subsequence also converges to $L$ . So if two subsequences of $\{a_n\}$ converge to different limits, then $\{a_n\}$ diverges.

In plain words When a sequence converges, every subsequence heads to the same limit. If the even terms head to one number and the odd terms to another, there is no single limit.

Example Suppose $(-1)^n \to L$ . With $\varepsilon = \frac{1}{2}$ there would be an $N$ such that both $\lvert 1 - L \rvert < \frac{1}{2}$ and $\lvert -1 - L \rvert < \frac{1}{2}$ , using an even and an odd $n > N$ . The triangle inequality would then give

2 = \lvert 1 - (-1) \rvert \le \lvert 1 - L \rvert + \lvert L - (-1) \rvert < \frac{1}{2} + \frac{1}{2} = 1,

which is false. So $\{(-1)^n\}$ diverges even though it is bounded.

Bounded monotonic sequences converge

Definition The monotone sequence theorem. Every bounded monotonic sequence converges. More precisely, if $\{a_n\}$ is nondecreasing and bounded above, then it converges to $\sup\{a_n\}$ , and if it is nonincreasing and bounded below, it converges to $\inf\{a_n\}$ .

In plain words A sequence that only goes up but has a ceiling has nowhere to go but to settle down. The theorem lets you prove a limit exists before you know its value.

Example Let $a_1 = 1$ and $a_{n+1} = \sqrt{2 + a_n}$ . If $0 < a_n < 2$ , then $0 < a_{n+1} < \sqrt{4} = 2$ , and $a_{n+1} > a_n$ because

a_{n+1}^2 - a_n^2 = 2 + a_n - a_n^2 = (2 - a_n)(1 + a_n) > 0.

By induction the sequence is increasing and bounded above by $2$ , so it converges to some $L$ . Letting $n \to \infty$ in the rule gives $L = \sqrt{2 + L}$ , so $L^2 - L - 2 = 0$ and $L = 2$ , since $L \ge 1$ .

$n$	$1$	$2$	$3$	$4$	$5$
$a_n$	$1$	$1.7321$	$1.9319$	$1.9829$	$1.9957$

Completeness

Definition The completeness axiom for the real numbers states that every nonempty set of real numbers that is bounded above has a least upper bound, its supremum, in $\mathbb{R}$ . This axiom is what makes the monotone sequence theorem true. Take a nondecreasing sequence that is bounded above, and let $L = \sup\{a_n\}$ . For each $\varepsilon > 0$ the number $L - \varepsilon$ is not an upper bound, so some $a_N > L - \varepsilon$ , and then

L - \varepsilon < a_N \le a_n \le L \quad \text{for all } n \ge N.

In plain words The real number line has no gaps. A sequence that climbs toward a ceiling always finds an actual number waiting at the top.

Example In the rational numbers alone this fails. The decimal truncations $1, 1.4, 1.41, 1.414, 1.4142, \dots$ of $\sqrt{2}$ are rational, nondecreasing and bounded above by $2$ . Their limit is $\sqrt{2}$ , which is not rational. So without the real numbers there would be no limit to converge to.

Check yourself

For $a_n = \frac{1}{n^2}$ and $\varepsilon = 0.001$ , find the smallest $N$ such that $\lvert a_n - 0 \rvert < \varepsilon$ for all $n > N$ .

Show the answer

The inequality $1/n^2 < 0.001$ means $n^2 > 1000$ , that is $n \ge 32$ since $31^2 = 961$ and $32^2 = 1024$ . So $N = 31$ .

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

Show the answer

No. The even terms tend to $1$ and the odd terms tend to $-1$ , so there is no single limit, even though the sequence is bounded.

Let $a_1 = 1$ and $a_{n+1} = \frac{a_n + 6}{2}$ . Show it converges and find the limit.

Show the answer

If $a_n < 6$ then $a_{n+1} < 6$ and $a_{n+1} - a_n = \frac{6 - a_n}{2} > 0$ , so by induction the sequence is increasing and bounded above by $6$ . So it converges to some $L$ , and letting $n \to \infty$ in the rule gives

L = \frac{L + 6}{2},

so $L = 6$ .

Convergence of Sequences

Key terms

The epsilon N definition

Finite limits

Divergence

Unboundedness

Oscillation

Bounded monotonic sequences converge

Completeness

Check yourself

How this lesson connects

Builds on

Used later in