Limit Superior and Limit Inferior¶

Definition¶

Let $(a_n)_{n=0}^{\infty}$ be a sequence. For each $N\in\mathbb{N}\text{,}$ consider the supremum and infimum of the terms from $a_N$ onward: $\displaystyle s_N = \sup_{n\geq N} a_n$ and $\displaystyle i_N = \inf_{n\geq N} a_n\text{.}$ The limit superior and limit inferior of $(a_n)_{n=0}^{\infty}$ are defined as:

\begin{aligned} \limsup_{n\to \infty}a_n &:= \lim_{N\to\infty} s_N = \inf_{N\geq 0} s_N\\ &=\inf_{N\geq 0}\sup_{n\geq N} a_n, \\ \liminf_{n\to \infty}a_n &:= \lim_{N\to\infty} i_N = \sup_{N\geq 0} i_N\\ &=\sup_{N\geq 0}\inf_{n\geq N} a_n. \end{aligned}

Apply the definition of a Limit of a Sequence to the tail suprema $s_N$ and infima $i_N$ . Their monotonicity ensures extended-real limits; if the original sequence has an ordinary limit, both tail limits equal it.

Interpretation¶

The limit superior and limit inferior describe where a sequence ends up in the long run, also when it does not converge: they are the eventual upper and lower bounds of its terms. Only the tail counts, so changing finitely many terms changes neither of them.

An oscillating sequence whose even terms lie above 1 and odd terms below minus 1 and approach these values, with step lines of the tail suprema decreasing towards 1 and the tail infima increasing towards minus 1.

The figure shows $a_n=(-1)^n(1+\tfrac{2}{n})$ $\frac{2}{n})$ . The tail suprema $s_N$ form the upper step line and decrease towards $1$ ; the tail infima $i_N$ form the lower one and increase towards $-1$ . So $\limsup a_n=1$ and $\liminf a_n=-1$ $- 1$ , although no term equals either value: the even terms $1+\tfrac{2}{n}$ lie above $1$ and the odd terms lie below $-1$ .

Move $N$ : $s_N=\sup_{n\ge N}a_n$ can only decrease and $i_N=\inf_{n\ge N}a_n$ can only increase. Their limits are $\limsup a_n$ and $\liminf a_n$ .

The tail supremum $s_N$ is taken over fewer terms as $N$ grows, so it can only decrease, and a monotone sequence always has a limit in the extended reals. For a bounded sequence there are two equivalent readings. First, $\limsup a_n$ is the largest limit of a convergent subsequence, and $\liminf a_n$ is the smallest. Second, if $L=\limsup a_n$ $lim sup a_{n}$ , then for every $\varepsilon>0$ the terms satisfy $a_n<L+\varepsilon$ for all large $n$ , and $a_n>L-\varepsilon$ for infinitely many $n$ .

These quantities appear wherever a sequence of interest need not converge. The root test uses $\limsup\sqrt[n]{|a_n|}$ , and the radius of convergence of a power series is $1/\limsup|c_n|^{1/n}$ .

Two slips are common. The limit superior is not the supremum of the sequence: above, $\sup_n a_n=a_2=2$ $2$ , while $\limsup a_n=1$ $1$ . And it does not behave like a limit under addition: for $a_n=(-1)^n$ and $b_n=-(-1)^n$ $- (- 1)^{n}$ , $\limsup(a_n+b_n)=0$ $0$ , while $\limsup a_n+\limsup b_n=2$ $2$ . In general only $\limsup(a_n+b_n)\le\limsup a_n+\limsup b_n$ holds, when the right-hand side is defined.

Key Results¶

The limit superior and limit inferior of a real sequence always exist in the extended real numbers $[-\infty,+\infty]$ .
A sequence converges if and only if its $\limsup$ and $\liminf$ are equal and finite; otherwise, it's divergent.

Examples¶

For $a_n = (-1)^n$ $(- 1)^{n}$ : $\limsup\limits_{n\to \infty}a_n = 1\text{,}$ $\liminf\limits_{n\to \infty}a_n = -1\text{.}$
For $a_n = 1/2^n$ $1/ 2^{n}$ : $\limsup\limits_{n\to \infty}a_n = \liminf\limits_{n\to \infty}a_n = 0\text{.}$

Prerequisites¶

Accumulation Point in a Metric Space

Sources¶

Math III: Analysis · Lecture 2

Aliases: lim sup, lim inf, upper limit, lower limit, limsup, liminf Symbols: element of, epsilon, infinity, natural numbers, square root

Last updated: 20 October 2026 03:02