Convergence and the Two Tail Limits
The limsup and liminf describe the eventual upper and lower behavior of a sequence. The previous tutorials developed their properties separately. We now connect both quantities to ordinary convergence: when a sequence converges to a finite real number, its upper and lower tail limits must agree there. Conversely, if both tail limits agree at a finite value, the sequence must converge to that value.
For a real sequence \((a_n)\), write
The limsup is the limit of the nonincreasing tail suprema, and the liminf is the limit of the nondecreasing tail infima, with extended-real values allowed:
A tail supremum or infimum need not be attained by a term of the sequence. That does not prevent it from describing the tail's upper or lower bound. In particular, a tail can have supremum \(L\) even when every term in that tail is strictly less than \(L\).
A Convergent Sequence Has Matching Tail Limits
Suppose \(a_n\to L\in\mathbb{R}\). Given any positive error, every sufficiently late term lies within that error of \(L\). Consequently, all sufficiently late tails are squeezed into the same small interval. Their suprema and infima must lie there too, even if those bounds are not attained.
Proof. A convergent sequence is bounded, so each of its tails has a finite supremum and infimum. Fix \(\varepsilon>0\). By convergence, there is an \(N_0\) such that $$ L-\varepsilon<a_n<L+\varepsilon \qquad(n\geq N_0). $$ For every \(N\geq N_0\), the terms in the \(N\)-th tail satisfy these same inequalities. Therefore $$ L-\varepsilon\leq i_N(a)\leq s_N(a)\leq L+\varepsilon. $$ The tail suprema decrease with \(N\), and the tail infima increase with \(N\). Since both lie between \(L-\varepsilon\) and \(L+\varepsilon\) for every \(N\geq N_0\), their limits lie in that interval as well. This holds for every \(\varepsilon>0\), so both limits equal \(L\). By the definitions of limsup and liminf, the two claimed identities follow. \(\square\)
The weak inequalities at the endpoints matter: the supremum of numbers strictly below \(L+\varepsilon\) may equal \(L+\varepsilon\), and the infimum of numbers strictly above \(L-\varepsilon\) may equal \(L-\varepsilon\). The argument does not assume either endpoint is attained.
Worked Example: A Decreasing Sequence Approaches Its Limsup
For \(n\in\mathbb{N}_0\), let $$ a_n=5+\frac{1}{n+2}. $$ The sequence is decreasing, so the supremum of the tail beginning at \(N\) is its first term: $$ s_N(a)=5+\frac{1}{N+2}. $$ Every term in that tail is greater than \(5\), and the terms approach \(5\) as the indices increase. Thus \(i_N(a)=5\): it is the infimum, although no term equals it. It follows that $$ \limsup_{n\to\infty}a_n =\inf_N\left(5+\frac{1}{N+2}\right)=5, \qquad \liminf_{n\to\infty}a_n =\sup_N 5=5. $$ The ordinary limit is \(5\), in agreement with the theorem.
Worked Example: A Convergent Sequence That Oscillates
Define $$ a_n=-2+\frac{(-1)^n}{n+1}. $$ The sign of the error alternates, but its magnitude satisfies $$ |a_n-(-2)|=\frac{1}{n+1}. $$ For every \(n\geq N\), \(1/(n+1)\leq1/(N+1)\). Hence every term of the \(N\)-th tail lies between \(-2-1/(N+1)\) and \(-2+1/(N+1)\), and therefore $$ -2-\frac{1}{N+1}\leq i_N(a)\leq s_N(a)\leq -2+\frac{1}{N+1}. $$ As \(N\) increases, both bounds approach \(-2\), so the tail infima and tail suprema both approach \(-2\). Thus $$ \liminf_{n\to\infty}a_n=\limsup_{n\to\infty}a_n=-2. $$ Alternation does not prevent convergence when the size of the oscillation tends to zero.
Equality at a Finite Value Forces Convergence
The converse gives a useful test. If the limsup is a finite number \(L\), the Threshold Characterization of a Finite Limsup says that the sequence is eventually below \(L+\varepsilon\), for every \(\varepsilon>0\). If the liminf is also \(L\), its Threshold Characterization says the sequence is eventually above \(L-\varepsilon\). Together these eventual bounds place the sequence in every neighborhood of \(L\).
Proof. The forward direction is the theorem just proved. For the reverse direction, suppose both the limsup and liminf equal the finite real number \(L\). Fix \(\varepsilon>0\). By the Threshold Characterization of a Finite Limsup, there is an \(N_1\) such that $$ a_n<L+\varepsilon \qquad(n\geq N_1). $$ By the Threshold Characterization of a Finite Liminf, there is an \(N_2\) such that $$ a_n>L-\varepsilon \qquad(n\geq N_2). $$ For \(n\geq\max\{N_1,N_2\}\), both inequalities hold, giving $$ L-\varepsilon<a_n<L+\varepsilon. $$ Equivalently, \(|a_n-L|<\varepsilon\) for every sufficiently large \(n\). This is the definition of \(a_n\to L\). \(\square\)
Worked Example: Persistent Oscillation Produces Different Tail Limits
Let \(a_n=4+(-1)^n\). If \(n\) is even, then \(a_n=4+1=5\); if \(n\) is odd, then \(a_n=4-1=3\). Every tail contains even and odd indices, so every tail supremum is \(5\) and every tail infimum is \(3\). Consequently, $$ \limsup_{n\to\infty}a_n=5, \qquad \liminf_{n\to\infty}a_n=3. $$ The values \(5\) and \(3\) both occur arbitrarily far out, so the sequence cannot settle near a single finite number. The two tail limits record the persistent spread, rather than a common value.
Why Finiteness Matters
The criterion requires the shared value to be finite. Limsup and liminf are allowed to take the extended-real values \(+\infty\) and \(-\infty\), but those values are not ordinary finite limits. In fact, a sequence can have equal limsup and liminf at \(+\infty\) while diverging upward.
Worked Example: Equal Infinite Tail Limits Do Not Give a Finite Limit
Take \(a_n=n+1\) for \(n\in\mathbb{N}_0\). The sequence is increasing and unbounded above. The infimum of the tail starting at \(N\) is its first term, \(i_N(a)=N+1\), while every tail is unbounded above, so \(s_N(a)=+\infty\). Hence $$ \liminf_{n\to\infty}a_n=\sup_N(N+1)=+\infty, \qquad \limsup_{n\to\infty}a_n=\inf_N(+\infty)=+\infty. $$ The two extended-real values agree, but the sequence does not converge to a finite real number. It instead tends to \(+\infty\).
The finite equality criterion is therefore a statement about ordinary convergence, not a claim that every pair of equal extended-real tail limits gives a finite limit. When applying it, first check that the common value is a real number. The Threshold Characterizations used in its proof also have finite-value hypotheses, which is why the proof does not extend unchanged to an infinite common value.
Using Both Limits Together
For a finite limit, the limsup and liminf provide two complementary checks. The limsup controls how large terms can remain in late tails; the liminf controls how small they can remain. If both settle at the same finite number, no terms can persist outside an arbitrarily small interval around that number. If they differ, the sequence cannot converge to a finite limit. Examples with fixed-amplitude oscillation show one way they can differ; other nonconvergent sequences need not alternate between just two values.
A common pitfall is to treat a finite initial exception as evidence against the criterion. Limsup and liminf depend on the behavior of tails as their starting index increases, so changing finitely many terms does not change either value. Another is to assume that the tail supremum or infimum must be an actual term. The first worked example shows why this is unnecessary: a tail can approach its infimum without ever reaching it. The criterion concerns the bounds of increasingly late tails, not whether those bounds occur among the terms.
Check Your Understanding
Use the tail definitions and the results proved here to answer the following questions.
- If \(a_n\to L\in\mathbb{R}\), what are the limsup and liminf of \((a_n)\)?
- Why can a tail supremum equal an endpoint even if every term in the tail is strictly below it?
- Which two eventual bounds establish convergence when the limsup and liminf both equal a finite number \(L\)?
- For \(a_n=4+(-1)^n\), what are the tail supremum and tail infimum?
- Why does equality of limsup and liminf at \(+\infty\) not imply convergence to a finite real number?