Why Look at the Tails?
A sequence can oscillate without converging, or it can have terms that grow without bound in one direction while repeatedly returning to a finite range. Its behavior near the beginning may not reflect what happens far out. To describe its long-term upper behavior, we examine the suprema of its tails: the sets of terms from a given index onward. As the starting index moves forward, each tail becomes smaller, so its supremum cannot increase.
For a bounded sequence, the upper limit has already appeared as a subsequential limit and as the limiting value of the tail suprema. We now define the limsup for every real sequence, including sequences that are unbounded. This makes it possible to distinguish three cases: a finite upper limiting value, tails that remain unbounded above, and sequences that eventually lie below every real bound.
Definition by Tail Suprema
Each tail is nonempty, so its supremum is either a real number or \(+\infty\), never \(-\infty\). Since the tail beginning at \(N+1\) is contained in the tail beginning at \(N\), we have \(s_{N+1}\leq s_N\). Thus the tail suprema form a nonincreasing sequence in the extended sense. Their infimum records the level toward which they descend, including the possibilities of descending without bound or remaining infinite.
The definition does not require the supremum of a tail to be one of its terms. For example, a tail may contain terms that approach its supremum without attaining it. Nor does the limsup require the original sequence to be bounded below. It is the upper behavior of the tails that matters.
Finite Limsup as an Eventual Upper Threshold
Suppose the limsup is a finite real number \(L\). Then every number strictly above \(L\) eventually bounds all terms, while every number strictly below \(L\) is exceeded arbitrarily far out. This gives \(L\) a useful interpretation: it is the least eventual upper threshold.
- For every \(\varepsilon>0\), there is an \(N\) such that \(a_n<L+\varepsilon\) for every \(n\geq N\).
- For every \(\varepsilon>0\) and every \(N\), there is an \(n\geq N\) such that \(a_n>L-\varepsilon\).
Proof. Suppose first that \(\inf_N s_N=L\). Given \(\varepsilon>0\), the definition of infimum gives an \(N\) with \(s_N<L+\varepsilon\). Every term in that tail satisfies \(a_n\leq s_N<L+\varepsilon\), proving the first property.
For the second property, fix \(\varepsilon>0\) and \(N\). Since \(L\) is the infimum of all the \(s_j\), we have \(s_N\geq L>L-\varepsilon\). If every \(n\geq N\) satisfied \(a_n\leq L-\varepsilon\), then \(L-\varepsilon\) would be an upper bound for that tail, giving \(s_N\leq L-\varepsilon\). This contradicts \(s_N>L-\varepsilon\). Hence some \(n\geq N\) satisfies \(a_n>L-\varepsilon\).
Conversely, suppose the two stated properties hold. The first implies that for every \(\varepsilon>0\), some tail has supremum at most \(L+\varepsilon\). Therefore \(\inf_N s_N\leq L+\varepsilon\) for every \(\varepsilon>0\). The second implies that every tail contains a term greater than \(L-\varepsilon\), so every \(s_N\geq L-\varepsilon\), and consequently \(\inf_N s_N\geq L-\varepsilon\). Since these inequalities hold for every positive \(\varepsilon\), the infimum equals \(L\). \(\square\)
Worked Example: A Finite Limsup Despite Unbounded Terms
Define a sequence by $$ a_n= \begin{cases} 1+\dfrac{1}{n+1},&n\text{ even},\\[4pt] -n,&n\text{ odd}. \end{cases} $$ The odd-indexed terms are unbounded below, but they do not affect the upper tail behavior once an even-indexed term is present. In any tail beginning at \(N\), let \(m\) be the least even integer with \(m\geq N\). The even-indexed terms in that tail are \(1+1/(n+1)\) for even \(n\geq m\); these decrease as \(n\) increases, so their largest value is \(1+1/(m+1)\). The odd-indexed terms are negative, whereas \(1+1/(m+1)>1\). Thus $$ s_N=1+\frac{1}{m+1}. $$ As \(N\to\infty\), its least even integer \(m\geq N\) also tends to infinity, and \(s_N\to1\). Therefore \(\limsup a_n=1\). The even-indexed subsequence approaches \(1\), while the sequence as a whole is unbounded below.
A Subsequence Approaching a Finite Limsup
The threshold characterization also allows us to select terms that approach the limsup. The lower-threshold property supplies terms above \(L-\varepsilon\) arbitrarily far out; the eventual upper bound keeps sufficiently late terms below \(L+\varepsilon\). Choosing the terms successively farther out gives convergence.
Proof. For each positive integer \(k\), apply the first part of the Threshold Characterization with \(\varepsilon=1/(k+1)\). Choose \(N_k\) such that $$ a_n<L+\frac{1}{k+1}\qquad(n\geq N_k). $$ We select indices \(n_k\) recursively. By the second part of the characterization with \(\varepsilon=1/2\) and \(N=N_1\), choose \(n_1\geq N_1\) such that \(a_{n_1}>L-1/2\). Given \(n_{k-1}\), use the second part of the characterization with \(\varepsilon=1/(k+1)\) and \(N=\max\{N_k,n_{k-1}+1\}\). It gives an index \(n_k\geq\max\{N_k,n_{k-1}+1\}\) such that $$ a_{n_k}>L-\frac{1}{k+1}. $$ Because \(n_k\geq N_k\), the upper bound also gives \(a_{n_k}<L+1/(k+1)\). Therefore $$ \left|a_{n_k}-L\right|<\frac{1}{k+1}. $$ The right-hand side tends to zero, so \(a_{n_k}\to L\). The indices strictly increase by construction, making this a subsequence. \(\square\)
The proof above does not require boundedness. The result here also applies when the sequence is unbounded below, as the preceding example illustrates. The finite limsup controls the upper tail even if there is no lower bound on the sequence.
Worked Example: Tail Suprema That Are Never Attained
Let \(a_n=6-1/(n+2)\) for \(n\in\mathbb{N}_0\). The terms increase toward \(6\), because $$ a_{n+1}-a_n =\left(6-\frac{1}{n+3}\right)-\left(6-\frac{1}{n+2}\right) =\frac{1}{n+2}-\frac{1}{n+3} =\frac{1}{(n+2)(n+3)}>0. $$ Every term is less than \(6\), and the terms approach \(6\). In any tail, the terms therefore have supremum \(6\), even though no term equals \(6\). Thus \(s_N=6\) for every \(N\), and \(\limsup a_n=6\). Here the tail suprema are constant, but the sequence itself converges to their value.
The Two Infinite Cases
The extended values of the limsup have direct interpretations. If the limsup is \(+\infty\), every tail is unbounded above. If it is \(-\infty\), the sequence tends to \(-\infty\). These cases are different: the first describes repeated arbitrarily large values no matter how far out we go, while the second says all sufficiently late terms lie below any specified real threshold.
- \(\limsup_{n\to\infty}a_n=+\infty\) if and only if every tail \(\{a_n:n\geq N\}\) is unbounded above.
- \(\limsup_{n\to\infty}a_n=-\infty\) if and only if \(a_n\to-\infty\).
Proof. If the infimum of the tail suprema is \(+\infty\), then every \(s_N=+\infty\), because each \(s_N\) is at least that infimum. Thus every tail is unbounded above. Conversely, if every tail is unbounded above, then every \(s_N=+\infty\), so their infimum is \(+\infty\).
Now suppose \(\inf_N s_N=-\infty\). Given any real \(A\), there is an \(N\) with \(s_N<A\); otherwise every \(s_N\geq A\), which would force their infimum to be at least \(A\). For every \(n\geq N\), we then have \(a_n\leq s_N<A\). Since this holds for every real \(A\), the sequence tends to \(-\infty\).
Conversely, suppose \(a_n\to-\infty\). For any real \(A\), there is an \(N\) such that \(a_n<A\) for every \(n\geq N\). Hence \(s_N\leq A\), and therefore \(\inf_j s_j\leq A\). This is true for every real \(A\), so the infimum of the tail suprema is \(-\infty\). \(\square\)
Worked Example: Limsup Equal to Positive Infinity
Let \(a_n=n^2\). Fix any starting index \(N\) and any real number \(C\). Choose an integer \(n\geq N\) with \(n^2>C\); such an integer exists because the nonnegative integers are unbounded. Thus every tail is unbounded above, and the Infinite Limsup Criteria give \(\limsup a_n=+\infty\).
A sequence need not tend to \(+\infty\) for its limsup to be \(+\infty\): the criterion requires unboundedness above in every tail, not that all sufficiently late terms exceed each fixed threshold. For example, \(a_n=n\) when \(n\) is even and \(a_n=0\) when \(n\) is odd has an unbounded-above tail after every index, but its odd-indexed terms remain zero.
Worked Example: Limsup Equal to Negative Infinity
Let \(a_n=4-\sqrt{n+1}\). Given any real \(A\), choose \(N\) so large that \(\sqrt{N+1}>4-A\) when \(4-A\geq0\); if \(4-A<0\), every term already satisfies \(4-\sqrt{n+1}<A\) for sufficiently large \(n\), and in fact \(4-\sqrt{n+1}\leq4<A\) for all \(n\). In the first case, for every \(n\geq N\), $$ a_n=4-\sqrt{n+1}\leq4-\sqrt{N+1}<A. $$ Thus \(a_n\to-\infty\), and the Infinite Limsup Criteria imply \(\limsup a_n=-\infty\).
Using the Limsup Carefully
The limsup is not necessarily the largest value taken by the sequence. It is a limiting upper threshold for the tails. A sequence may never attain its limsup, as in the sequence \(6-1/(n+2)\), or may attain it infinitely often. Conversely, a large term among the first few entries has no effect on the limsup: deleting or changing finitely many terms leaves all sufficiently late tails unchanged.
For a finite limsup \(L\), the two threshold conditions work together. The eventual upper bound prevents terms from staying above \(L+\varepsilon\), while the repeated exceedance of \(L-\varepsilon\) prevents the tails from settling strictly below \(L\). The subsequence theorem packages both facts into a sequence of terms converging to \(L\). When the limsup is infinite, use the corresponding tail criterion instead; do not treat \(+\infty\) or \(-\infty\) as ordinary real subsequential limits.
Check Your Understanding
Use the tail-supremum definition and the proved criteria to answer the following questions.
- Why are the tail suprema \(s_N\) nonincreasing as \(N\) increases?
- If a sequence has finite limsup \(L\), what happens eventually to its terms relative to \(L+\varepsilon\), and what must happen relative to \(L-\varepsilon\) in every tail?
- Can a finite limsup be the supremum of every tail without being attained by any term? Explain using the example \(a_n=6-1/(n+2)\).
- What condition on every tail characterizes limsup equal to \(+\infty\)?
- Why does \(a_n\to-\infty\) imply that the limsup is \(-\infty\)?