limsup and liminf for Bounded Sequences

Bounds, Cluster Points, and Tail Behavior

For a bounded sequence, the limsup and liminf are finite numbers that describe the upper and lower behavior of its late tails. The previous tutorial showed that convergence occurs exactly when these two values agree at a finite number. Here we look more closely at what they record when they do not agree: they locate the extreme cluster points and measure how much variation remains in the tails.

For a real sequence $(a_n)$, write

s_N(a)=\sup\{a_n:n\geq N\}, \qquad i_N(a)=\inf\{a_n:n\geq N\}.

If $(a_n)$ is bounded, each tail has a finite supremum and infimum. As $N$ increases, the tail suprema are nonincreasing and the tail infima are nondecreasing. Their limits are the limsup and liminf:

U=\limsup_{n\to\infty}a_n=\inf_{N\in\mathbb{N}_0}s_N(a), \qquad I=\liminf_{n\to\infty}a_n=\sup_{N\in\mathbb{N}_0}i_N(a).

Because every tail is nonempty and its infimum is at most its supremum, $i_N(a)\leq s_N(a)$ for every $N$. For any $N,M$, let $K=\max\{N,M\}$. The tail infima are nondecreasing and the tail suprema are nonincreasing, so $i_N(a)\leq i_K(a)\leq s_K(a)\leq s_M(a)$. Taking the supremum over $N$ and then the infimum over $M$ gives $I\leq U$. Unlike for an unbounded sequence, neither value here is infinite.

Definition: A real number $x$ is a cluster point of $(a_n)$ if some subsequence of $(a_n)$ converges to $x$. The cluster-point set consists of all such real numbers.

The Extreme Cluster Points

For bounded sequences, the limsup and liminf do more than bound the cluster points: they are themselves cluster points, and they are the greatest and least ones. The existence of subsequences converging to the limsup and liminf is the result established in “Bolzano-Weierstrass and Bounded Sequences.” The Threshold Characterizations of a Finite Limsup and a Finite Liminf also describe the eventual bounds around these values. Together these results give a useful picture of the entire cluster-point set.

Theorem (Cluster Points of a Bounded Sequence): Let $(a_n)$ be bounded, with $I=\liminf_{n\to\infty}a_n$ and $U=\limsup_{n\to\infty}a_n$. Every cluster point $x$ of $(a_n)$ satisfies $I\leq x\leq U$. Both $I$ and $U$ are cluster points. In particular, $I$ is the least cluster point and $U$ is the greatest cluster point.

Proof. Theorem (Upper and Lower Limits Are Subsequential Limits), established earlier in “Bolzano-Weierstrass and Bounded Sequences,” gives a subsequence converging to $I$ and a subsequence converging to $U$. Thus both endpoints are cluster points.

Now let $x$ be any cluster point, and choose a subsequence $(a_{n_k})$ converging to $x$. Fix $\varepsilon>0$. By the Threshold Characterization of a Finite Limsup, there is an $N_1$ such that

a_n<U+\varepsilon \qquad(n\geq N_1).

By the Threshold Characterization of a Finite Liminf, there is an $N_2$ such that

a_n>I-\varepsilon \qquad(n\geq N_2).

The indices $n_k$ increase without bound, so all sufficiently late terms of the subsequence satisfy both inequalities. Passing to the limit along that subsequence gives $x\leq U+\varepsilon$ and $x\geq I-\varepsilon$. Since this holds for every $\varepsilon>0$, we obtain $I\leq x\leq U$. As $I$ and $U$ are themselves cluster points, they are respectively the least and greatest cluster points. $\square$

The theorem does not say that every number between $I$ and $U$ is a cluster point. The cluster-point set can have gaps. What it does say is that no subsequence can converge outside the interval $[I,U]$, and that subsequences do approach both ends of this interval.

Worked Example: Two Cluster Points with a Gap Between Them

Let $a_n=2(-1)^n$ for $n\in\mathbb{N}_0$. Even indices give $a_n=2$, while odd indices give $a_n=-2$. Every tail contains both an even and an odd index, and all its terms are either $-2$ or $2$. Consequently,

s_N(a)=2,\qquad i_N(a)=-2 \qquad\text{for every }N.

Thus $U=2$ and $I=-2$. The even-indexed subsequence is constantly $2$, and the odd-indexed subsequence is constantly $-2$, so both endpoints are cluster points. There are no other cluster points: any convergent subsequence must have infinitely many terms drawn from at least one of the two values, and a subsequence containing infinitely many of both cannot converge because its terms keep taking values a distance $4$ apart. The cluster-point set is exactly $\{-2,2\}$, not the whole interval $[-2,2]$.

Tail Oscillation

The difference between the supremum and infimum of a tail is its diameter: it measures the largest spread that terms in that tail can have, allowing for a supremum or infimum that is not attained. Define the tail oscillation by

$$ D_N=s_N(a)-i_N(a). $$

For a bounded sequence, $D_N\geq0$. It cannot increase as $N$ increases: removing early terms from a tail cannot increase its supremum or decrease its infimum. The limiting value of $D_N$ is exactly the gap between the limsup and the liminf.

Theorem (Limiting Tail Oscillation): If $(a_n)$ is bounded, then $$ \lim_{N\to\infty}\bigl(s_N(a)-i_N(a)\bigr) = \limsup_{n\to\infty}a_n-\liminf_{n\to\infty}a_n. $$

Proof. Write $U=\limsup a_n$ and $I=\liminf a_n$. Since $U$ is the infimum of the tail suprema, $s_N(a)\geq U$ for every $N$. Since $I$ is the supremum of the tail infima, $i_N(a)\leq I$ for every $N$. Hence

D_N=s_N(a)-i_N(a)\geq U-I.

Fix $\varepsilon>0$. The nonincreasing sequence $s_N(a)$ converges to $U$, so for all sufficiently large $N$, $s_N(a)<U+\varepsilon/2$. The nondecreasing sequence $i_N(a)$ converges to $I$, so for all sufficiently large $N$, $i_N(a)>I-\varepsilon/2$. Taking $N$ large enough for both inequalities gives

D_N=s_N(a)-i_N(a) < \left(U+\frac{\varepsilon}{2}\right) - \left(I-\frac{\varepsilon}{2}\right) =U-I+\varepsilon.

Thus, for every $\varepsilon>0$, all sufficiently late $D_N$ lie between $U-I$ and $U-I+\varepsilon$. This proves that $D_N\to U-I$. $\square$

The limit $U-I$ describes the oscillation that persists arbitrarily far out. A tail can have positive diameter even when neither extreme is attained in that tail. The formula uses the supremum and infimum precisely so it still applies in that situation.

Worked Example: Alternating Terms Approach Two Unattained Bounds

For $n\in\mathbb{N}_0$, let

a_n=3+(-1)^n\left(1-\frac{1}{n+2}\right).

The factor $1-\frac{1}{n+2}$ is positive and increases toward $1$. At even indices, the terms are

a_n=4-\frac{1}{n+2}<4,

and at odd indices, they are

a_n=2+\frac{1}{n+2}>2.

In every tail, the even-indexed terms approach $4$ from below and the odd-indexed terms approach $2$ from above. All terms lie strictly between $2$ and $4$. Therefore each tail has supremum $4$ and infimum $2$, although neither bound is attained, so $s_N(a)=4$, $i_N(a)=2$, $U=4$, and $I=2$. The tail oscillation is $D_N=4-2=2$ for every $N$, agreeing with $U-I=2$.

Two Different Cluster-Point Patterns

The limsup and liminf summarize the extremes but do not, by themselves, describe how many cluster points lie between them. A bounded sequence may have just two cluster points, as in the alternating example, or it may have every point of an interval as a cluster point. The next construction verifies the latter possibility directly.

Worked Example: Every Point in an Interval Is a Cluster Point

Form a sequence by concatenating the finite blocks

B_k=\left(0,\frac{1}{k},\frac{2}{k},\ldots,\frac{k}{k}\right), \qquad k=1,2,3,\ldots.

Every term lies in $[0,1]$, and every block contains both $0$ and $1$. Every tail therefore contains $0$ and $1$ from some later complete block, while no term lies outside $[0,1]$. It follows that $i_N(a)=0$ and $s_N(a)=1$ for every $N$, so $I=0$ and $U=1$.

Now fix $x\in[0,1]$. In block $B_k$, choose the term $j_k/k$, where $j_k=\lfloor kx\rfloor$. For $x=1$, this gives $j_k=k$ and $j_k/k=1$. For $0\leq x<1$, the defining property of the floor gives $j_k\leq kx<j_k+1$, and therefore

0\leq x-\frac{j_k}{k}<\frac{1}{k}.

In either case, the selected term in block $B_k$ converges to $x$ as $k\to\infty$. Since these selected terms occur in successive blocks, they form a subsequence. Thus every $x\in[0,1]$ is a cluster point. The theorem on cluster points of bounded sequences shows that no cluster point can lie outside $[0,1]$, so the cluster-point set is exactly $[0,1]$.

How the Two Numbers Can Be Used

For a bounded sequence, $I$ and $U$ answer complementary questions. The limsup is the greatest value that subsequences can approach, and the liminf is the least. Their difference is the limiting tail oscillation. These statements help separate three ideas that can otherwise be conflated: bounds on individual terms, subsequential limits, and the amount of variation left in late tails.

A common pitfall is to infer that every number between the liminf and limsup must be a subsequential limit. The sequence taking only the values $-2$ and $2$ disproves this: its extremes are $-2$ and $2$, but intermediate values are not cluster points. Another is to assume the supremum or infimum of a tail must be one of its terms. The alternating sequence approaching $2$ and $4$ shows that both tail bounds can remain unattained.

Finally, the limiting tail oscillation connects this discussion to the Tail-Diameter Characterization from “Cauchy Criterion for Real Sequences.” For a bounded sequence, $D_N\to0$ exactly when the sequence is Cauchy. By the Cauchy Criterion for Real Sequences, that is exactly when the sequence converges to a finite limit. The formula $D_N\to U-I$ therefore explains why a positive gap between the liminf and limsup records persistent variation, while a zero gap corresponds to vanishing tail diameter.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

Why must the limsup and liminf of a bounded sequence be finite?
What bounds does the liminf and limsup place on any cluster point?
Can every number between the liminf and limsup be a cluster point? Give an example that supports your answer.
What is the limit of the tail oscillations $s_N(a)-i_N(a)$ for a bounded sequence?
In the block construction, why does selecting one term from each block produce a subsequence converging to any chosen $x\in[0,1]$?