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 limits, as $N\to\infty$ $\infty$ , of the supremum and the infimum of the tail $a_N,a_{N+1},\ldots$ . 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 $a_n=(-1)^n(1+\tfrac{2}{n})$ for $n\ge1$ $1$ . 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 $-1$ .

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: 27 October 2026 11:52