Bounded: A Sequence Inside a Fence

Is there a fence that pens in every term of the sequence at once? If yes, the sequence is bounded.

The fence may not move from term to term. A single number

M

must satisfy

|a_n| \le M

for every term. Every convergent sequence is bounded. But bounded does not imply convergent:

(-1)^n

just bounces back and forth inside its fence.

For example, a 🐑 sheep wanders around inside its fence. However far it roams, it never leaves the fence. A bounded sequence is the same picture.

Examples & Exercises

Show the sequence $a_n = \frac{n}{n + 1}$ is bounded, and prove completely that its supremum is 1, a value no term ever reaches.

0 < \frac{n}{n + 1} < 1, \qquad \sup_n a_n = 1

Approaching 1 but never touching it.

Show proof

Boundedness: for $n \ge 1$ , $0 < n < n + 1$ , so $0 < a_n < 1$ . Hence every term satisfies $|a_n| \le 1$ : the sequence is bounded.
1 is the least upper bound: given $\varepsilon > 0$ , the Archimedean property gives $n$ with $\frac{1}{\varepsilon} < n + 1$ . For that $n$ , $a_n = 1 - \frac{1}{n + 1} > 1 - \varepsilon$ . So no number below 1 is an upper bound.
Hence $\sup a_n = 1$ . And since every term satisfies $a_n < 1$ , no term attains the supremum.

Prove completely that every convergent sequence is bounded.

a_n \to L \ \Longrightarrow \ (a_n) \text{ is bounded}

A convergent example — the tail is trapped near the limit, and the head has only finitely many terms.

Show proof

Suppose $a_n \to L$ . Applying the definition of convergence with $\varepsilon = 1$ gives $N$ such that every $n > N$ has $|a_n - L| < 1$ , hence $|a_n| < |L| + 1$ . The tail is trapped.
The head is finite: $a_1, \dots, a_N$ are finitely many numbers, so $M_0 = \max(|a_1|, \dots, |a_N|)$ exists.
With $M = \max(M_0, |L| + 1)$ , both head terms and tail terms satisfy $|a_n| \le M$ . So the sequence is bounded.

Prove completely that the sequence $a_n = (-1)^n$ is bounded but does not converge.

|a_n| = 1 \ \forall n, \qquad a_n \not\to L \ \forall L

Inside the fence, just bouncing back and forth.

Show proof

Boundedness: every term has $|a_n| = 1$ , so $M = 1$ is a fence.
Non-convergence: suppose it converged to some real $L$ . Neighbouring terms satisfy $|a_{n + 1} - a_n| = 2$ . But convergence with $\varepsilon = \frac{1}{2}$ puts every term past some $N$ within $\frac{1}{2}$ of $L$ . Then neighbouring terms would satisfy $|a_{n + 1} - a_n| \le |a_{n + 1} - L| + |L - a_n| < 1$ by the triangle inequality.
But $2 < 1$ is impossible — a contradiction. So it converges to no $L$ . A bounded sequence need not converge.