Limit Superior & Inferior

An eventual ceiling and floor that exist even with no limit.

The idea

Every sequence has a limit superior and a limit inferior — an eventual ceiling and an eventual floor — even when it has no limit.

Fix an index $n$ and take the supremum of the tail from $n$ onward, $s_{n} = \sup_{k \ge n} a_{k}$, the highest value the sequence reaches from index $n$ on. Discarding terms can only lower a supremum, so $s_{1} \ge s_{2} \ge s_{3} \ge \cdots$. A decreasing sequence has a limit once we allow the value $-\infty$, and that limit is the limit superior: $\limsup_{n \to \infty} a_{n} = \lim_{n \to \infty} \ \sup_{k \ge n} a_{k}.$ The limit inferior $\liminf_{n \to \infty} a_{n}$ is the same construction with infima: its tail values $t_{n} = \inf_{k \ge n} a_{k}$ increase instead. Allowing $\pm\infty$, both quantities exist for every sequence, which the ordinary limit does not promise.

A second description is often quicker to compute with. A subsequential limit of $a_{n}$ is a value that some subsequence converges to. The limit superior is the largest subsequential limit, and the limit inferior is the smallest.

Always $\liminf a_{n} \le \limsup a_{n}$, and the sequence converges exactly when the two are equal and finite — their common value is then the limit.

Ways to work on it

• Walkthrough. Define limsup and liminf as the largest and smallest subsequential limits.
• Practice. Find the limsup or liminf of a non-convergent sequence.
• Hardest. A sequence with three cluster points bracketed by limsup and liminf.

Not sure where to start? Take the ten-question placement test.