Properties of limsup

From Tail Suprema to Limsup Properties

The limsup describes the long-term upper behavior of a sequence through the suprema of its tails. Its definition makes several useful properties visible: an eventual upper comparison between sequences gives a comparison of their limsups, and multiplying by a negative number reverses upper and lower behavior. We will also see how a convergent perturbation affects a limsup and why the limsup of a sum need not equal the sum of the individual limsups.

For each starting index $N$, the tail supremum is the least upper bound of all terms from that index onward. As the starting index increases, terms are removed from the tail, so its supremum cannot increase. The limsup is the infimum of these nonincreasing tail suprema.

Definition: For a real sequence $(a_n)$, define $$ s_N=\sup\{a_n:n\geq N\}, $$ where $s_N=+\infty$ if the tail is unbounded above. Then $$ \limsup_{n\to\infty}a_n=\inf_{N\in\mathbb{N}_0}s_N. $$ This infimum may be a finite real number or either of the extended values $-\infty$ and $+\infty$.

Since the $N+1$-st tail is contained in the $N$-th tail, $s_{N+1}\leq s_N$. A finite change to a sequence does not affect its limsup: sufficiently late tails of the two sequences are identical. More generally, taking the infimum over $N\geq N_0$, for any fixed $N_0$, gives the same limsup as taking it over all $N$. These observations are useful when the properties below involve only sufficiently late terms.

Eventual Comparison

If one sequence is eventually no larger than another, then no sufficiently late tail supremum of the first can exceed the corresponding tail supremum of the second. Passing from tail suprema to their infima gives the comparison of limsups.

Theorem (Order Preservation for Limsup): Suppose there is an $N_0\in\mathbb{N}_0$ such that $a_n\leq b_n$ for every $n\geq N_0$. Then $$ \limsup_{n\to\infty}a_n\leq\limsup_{n\to\infty}b_n. $$

Proof. For every $N\geq N_0$, each term in the $N$-th tail of $(a_n)$ is at most the corresponding term in the $N$-th tail of $(b_n)$. Thus $s_N(a)\leq s_N(b)$, where $s_N(a)$ and $s_N(b)$ denote their tail suprema. Taking infima over $N\geq N_0$ preserves this inequality. Since the tail suprema are nonincreasing, these infima over $N\geq N_0$ equal the infima over all $N$. The resulting inequality is exactly the claimed comparison of limsups. $\square$

Worked Example: Eventual Comparison Despite an Early Exception

Let $a_0=100$, and let $a_n=1/(n+1)$ for $n\geq1$. Let $b_n=2/(n+1)$ for every $n$. Although $a_0>b_0$, for every $n\geq1$, $$ a_n=\frac{1}{n+1}\leq\frac{2}{n+1}=b_n. $$ The tail beginning at $N\geq1$ has supremum $1/(N+1)$ for $(a_n)$ and $2/(N+1)$ for $(b_n)$, since both tails are decreasing. Both quantities tend to $0$. Therefore both limsups are $0$, in agreement with the eventual comparison theorem. The exceptional first term does not determine the limsup.

Multiplication by a Scalar

For a positive scalar, the order of the terms is preserved, so tail suprema scale by that scalar. A negative scalar reverses order: the supremum of the multiplied tail is the scalar times the infimum of the original tail. This explains why negative multiplication connects limsup to liminf rather than simply scaling limsup.

Theorem (Scalar Multiplication and Limsup): Let $c\in\mathbb{R}$. If $c>0$, then $$ \limsup_{n\to\infty}(ca_n)=c\limsup_{n\to\infty}a_n. $$ If $c<0$, then $$ \limsup_{n\to\infty}(ca_n)=c\liminf_{n\to\infty}a_n. $$ If $c=0$, then $\limsup_{n\to\infty}(ca_n)=0$. Products with extended values have their usual order-reversing meaning when $c<0$.

Proof. Write $s_N=\sup\{a_n:n\geq N\}$ and $i_N=\inf\{a_n:n\geq N\}$. If $c>0$, multiplication by $c$ preserves order, so the supremum of the multiplied tail is $cs_N$. Taking the infimum over $N$ gives $$ \limsup_{n\to\infty}(ca_n)=\inf_N(cs_N)=c\inf_Ns_N =c\limsup_{n\to\infty}a_n. $$ If $c<0$, multiplication reverses order, so the supremum of the multiplied tail is $ci_N$. Therefore $$ \limsup_{n\to\infty}(ca_n)=\inf_N(ci_N) =c\sup_N i_N =c\liminf_{n\to\infty}a_n. $$ The identities also hold when a tail is unbounded in the relevant direction, with the corresponding extended values. If $c=0$, every term $ca_n$ equals $0$, so its limsup is $0$. $\square$

Worked Example: A Negative Scalar Exchanges Upper and Lower Behavior

Define $a_n=3+(-1)^n$. When $n$ is even, $a_n=4$, and when $n$ is odd, $a_n=2$. Every tail contains both parities, so its supremum is $4$ and its infimum is $2$. Consequently, $$ \limsup a_n=4,\qquad \liminf a_n=2. $$ Take $c=-2$. The even-indexed terms of $(-2a_n)$ are $-8$, and the odd-indexed terms are $-4$; every tail therefore has supremum $-4$. Thus $\limsup(-2a_n)=-4$, which agrees with $$ (-2)\liminf a_n=(-2)(2)=-4. $$ Using the limsup $4$ instead would give $-8$, which is not the limsup of the multiplied sequence.

Adding a Convergent Sequence

A sequence that converges to $L$ eventually differs from the constant value $L$ by less than any chosen positive error. It follows that adding this sequence to $(a_n)$ changes the limsup by exactly $L$. The conclusion holds even when the limsup is infinite, because the perturbation stays bounded near its finite limit.

Theorem (Limsup Under a Convergent Perturbation): Suppose $b_n\to L\in\mathbb{R}$, and let $S=\limsup_{n\to\infty}a_n$. Then $$ \limsup_{n\to\infty}(a_n+b_n)=S+L. $$ Here $S+L$ has its usual extended-real meaning, since $L$ is finite.

Proof. First, for any fixed real number $c$, the supremum of the $N$-th tail of $(a_n+c)$ is $s_N+c$. Taking infima gives $$ \limsup_{n\to\infty}(a_n+c)=S+c. $$ Now fix $\varepsilon>0$. Since $b_n\to L$, there is an $N_0$ such that $|b_n-L|<\varepsilon$ whenever $n\geq N_0$. Hence $$ a_n+L-\varepsilon<a_n+b_n<a_n+L+\varepsilon \qquad(n\geq N_0). $$ By Order Preservation for Limsup and the constant-shift identity, $$ S+L-\varepsilon \leq\limsup_{n\to\infty}(a_n+b_n) \leq S+L+\varepsilon. $$ If $S$ is finite, these bounds for every $\varepsilon>0$ imply that the middle quantity equals $S+L$. If $S=+\infty$, the lower bound is $+\infty$, so the middle quantity is $+\infty$. If $S=-\infty$, the upper bound is $-\infty$, so the middle quantity is $-\infty$. This proves the theorem in all cases. $\square$

Worked Example: A Convergent Perturbation of an Unbounded Limsup

Let $a_n=n$ for even $n$, and $a_n=-n$ for odd $n$. Every tail contains even indices as large as desired, so every tail is unbounded above and $\limsup a_n=+\infty$. Let $b_n=5+1/(n+1)$. Since $1/(n+1)\to0$, we have $b_n\to5$. The convergent perturbation theorem gives $$ \limsup(a_n+b_n)=+\infty+5=+\infty. $$ Indeed, along even indices $a_n+b_n=n+5+1/(n+1)$, and these terms grow without bound. This example shows why the theorem does not require $(a_n)$ itself to be bounded.

A Bound for the Limsup of a Sum

When both individual limsups are finite, their sum provides an upper bound for the limsup of the termwise sum. Equality is not guaranteed: the terms of the two sequences may be large at different indices. The finite-limsup hypothesis is important for the stated result; unrestricted addition of extended limsups can involve an indeterminate expression such as $+\infty+(-\infty)$.

Theorem (Limsup Inequality for a Sum): Suppose $$ \limsup_{n\to\infty}a_n=A\in\mathbb{R}, \qquad \limsup_{n\to\infty}b_n=B\in\mathbb{R}. $$ Then $$ \limsup_{n\to\infty}(a_n+b_n)\leq A+B. $$

Proof. Fix $\varepsilon>0$. By the Threshold Characterization of a Finite Limsup, there are indices $N_a$ and $N_b$ such that $$ a_n<A+\varepsilon\quad(n\geq N_a), \qquad b_n<B+\varepsilon\quad(n\geq N_b). $$ For every $n\geq\max\{N_a,N_b\}$, adding these inequalities gives $$ a_n+b_n<A+B+2\varepsilon. $$ By Order Preservation for Limsup, comparison with the constant sequence $A+B+2\varepsilon$ yields $$ \limsup_{n\to\infty}(a_n+b_n)\leq A+B+2\varepsilon. $$ This holds for every $\varepsilon>0$. If the limsup on the left were greater than $A+B$, choosing $\varepsilon$ smaller than half the difference would contradict the displayed bound. Therefore it is at most $A+B$, as claimed. $\square$

Worked Example: The Sum Inequality Can Be Strict

Set $a_n=(-1)^n$ and $b_n=-(-1)^n$. Each sequence takes the values $1$ and $-1$ infinitely often, so every tail of each has supremum $1$. Thus $$ \limsup a_n=1,\qquad \limsup b_n=1. $$ But $a_n+b_n=0$ for every $n$, giving $$ \limsup(a_n+b_n)=0<2=\limsup a_n+\limsup b_n. $$ The terms achieve their upper values at opposite parities: when $a_n=1$, $b_n=-1$, and when $a_n=-1$, $b_n=1$. The two upper behaviors therefore do not occur together.

Using Limsup Properties Carefully

These properties answer different questions. Eventual comparison is useful for bounding a sequence above. Scalar multiplication by a negative number requires the liminf because the order reverses. A convergent perturbation shifts the limsup by a finite amount, while the sum inequality gives an upper bound rather than, in general, an equality.

A common mistake is to add limsups as though each sequence were simultaneously near its upper limiting value at the same indices. The strict-inequality example shows why that reasoning fails. Another is to apply a sum inequality without checking whether the extended values can be added meaningfully. Establish the relevant hypotheses first, then use the tail-supremum definition or a previously proved property that matches them.

Check Your Understanding

Use the tail-supremum definition and the proved results to answer the following questions.

Why can deleting finitely many terms of a sequence not change its limsup?
If $a_n\leq b_n$ for every sufficiently large $n$, what comparison follows for their limsups?
For $c<0$, why does $\limsup(ca_n)$ involve $\liminf a_n$ rather than $\limsup a_n$?
If $b_n\to L\in\mathbb{R}$, what is $\limsup(a_n+b_n)$ in terms of $\limsup a_n$, including when that limsup is infinite?
Why can $\limsup(a_n+b_n)$ be strictly less than $\limsup a_n+\limsup b_n$?

What You'll Learn