The liminf of a Sequence

Why Look at Tail Infima?

The limsup records the long-term upper behavior of a sequence by following the suprema of its tails. To describe the corresponding lower behavior, we follow the infima instead. This distinction matters for sequences that oscillate: their tails may keep producing terms near both an upper and a lower threshold, even when the sequence itself does not converge.

For a bounded sequence, the liminf has already appeared as a subsequential limit and as the limiting value of the tail infima. The definition below also applies to sequences unbounded above or below. Its extended values distinguish three possibilities: a finite lower limiting value, tails unbounded below, and terms that eventually exceed every real threshold.

Definition by Tail Infima

Definition: Let $(a_n)$ be a real sequence. For each $N\in\mathbb{N}_0$, define its $N$-th tail infimum by $$ i_N=\inf\{a_n:n\geq N\}, $$ where $i_N=-\infty$ if the tail is unbounded below. The liminf of $(a_n)$ is $$ \liminf_{n\to\infty}a_n=\sup_{N\in\mathbb{N}_0}i_N. $$ The supremum is allowed to be $-\infty$ or $+\infty$, so the liminf may take an extended real value.

Every tail is nonempty, so its infimum is either a real number or $-\infty$, never $+\infty$. As $N$ increases, the tail becomes smaller. Its infimum therefore cannot decrease: $i_{N+1}\geq i_N$. The liminf is the supremum of these nondecreasing tail infima.

The infimum of a tail need not be one of its terms. A tail may have terms that approach its infimum without attaining it. Also, the liminf is not necessarily the smallest value in the entire sequence: a finite number of early terms do not describe the behavior of increasingly late tails.

Finite Liminf as an Eventual Lower Threshold

Suppose the liminf is a finite real number $L$. Every number strictly below $L$ is eventually a lower bound for the sequence. On the other hand, no matter how far out we start, some term in that tail lies strictly below $L+\varepsilon$. Together, these properties identify $L$ as the greatest eventual lower threshold.

Theorem (Threshold Characterization of a Finite Liminf): If $\liminf_{n\to\infty}a_n=L\in\mathbb{R}$, then:

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 $a_n<L+\varepsilon$.

Conversely, if both properties hold for a real number $L$, then $\liminf_{n\to\infty}a_n=L$.

Proof. Write $i_N=\inf\{a_n:n\geq N\}$, so that $\sup_N i_N=L$. Given $\varepsilon>0$, the definition of supremum gives an $N$ such that $i_N>L-\varepsilon$. For every $n\geq N$, $$ a_n\geq i_N>L-\varepsilon. $$ This proves the eventual lower bound.

Now fix $\varepsilon>0$ and any $N$. Since $i_N\leq L<L+\varepsilon$, it cannot be the case that every $n\geq N$ satisfies $a_n\geq L+\varepsilon$. If that were so, $L+\varepsilon$ would be a lower bound for that tail, and $i_N\geq L+\varepsilon$, a contradiction. Thus some $n\geq N$ satisfies $a_n<L+\varepsilon$.

Conversely, suppose the two stated properties hold. The first property gives, for every $\varepsilon>0$, a tail whose infimum is at least $L-\varepsilon$. Hence $\sup_N i_N\geq L-\varepsilon$. The second property says that for every $N$, some term in the $N$-th tail is less than $L+\varepsilon$. Therefore $i_N\leq a_n<L+\varepsilon$ for that term, so $i_N<L+\varepsilon$ for every $N$, and $\sup_N i_N\leq L+\varepsilon$. Since these bounds hold for every $\varepsilon>0$, $\sup_N i_N=L$. $\square$

Worked Example: Finite Liminf with an Unbounded Sequence

Define $$ a_n= \begin{cases} -\dfrac{1}{n+1},&n\text{ even},\\[4pt] n+1,&n\text{ odd}. \end{cases} $$ The odd-indexed terms grow without bound, but they do not determine the lower behavior of a tail. In the tail beginning at $N$, let $m$ be the least even integer with $m\geq N$. The even-indexed terms in that tail are $-1/(n+1)$. They increase as the even index $n$ increases, so the least one is $-1/(m+1)$. Every odd-indexed term is positive. Consequently, $$ i_N=-\frac{1}{m+1}. $$ As $N$ increases, its least even integer $m\geq N$ also increases without bound, and $-1/(m+1)$ approaches $0$ from below. Thus $\sup_N i_N=0$, so $\liminf a_n=0$. The sequence is unbounded above, yet its liminf is finite.

A Subsequence Approaching a Finite Liminf

The threshold characterization can be used to select terms converging to a finite liminf. The first threshold property keeps all sufficiently late terms above $L-\varepsilon$. The second supplies a term below $L+\varepsilon$ in every tail. By choosing such terms successively farther out, we obtain a subsequence approaching $L$.

Theorem (Subsequence Approaching a Finite Liminf): If $\liminf_{n\to\infty}a_n=L\in\mathbb{R}$, then there is a subsequence of $(a_n)$ that converges to $L$.

Proof. For each positive integer $k$, the first part of the Threshold Characterization gives an index $N_k$ such that $$ a_n>L-\frac{1}{k+1}\qquad(n\geq N_k). $$ Choose $n_1\geq N_1$ using the second part of the characterization with $\varepsilon=1/2$, so that $a_{n_1}<L+1/2$. Once $n_{k-1}$ has been chosen, apply the second part of the characterization with $\varepsilon=1/(k+1)$ and with starting index $\max\{N_k,n_{k-1}+1\}$. It supplies an $n_k\geq\max\{N_k,n_{k-1}+1\}$ such that $$ a_{n_k}<L+\frac{1}{k+1}. $$ Since $n_k\geq N_k$, the lower 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, and hence the selected terms form a subsequence. $\square$

For bounded sequences, the earlier result “Upper and Lower Limits Are Subsequential Limits” already establishes the existence of subsequences approaching the liminf. The argument here also works when the sequence is unbounded above, as in the preceding example. A finite lower limiting value controls the lower behavior of the tails without requiring a global upper bound.

Worked Example: Tail Infima Need Not Be Attained

Let $a_n=2+1/(n+2)$. These terms are all greater than $2$, and $$ a_{n+1}-a_n =\frac{1}{n+3}-\frac{1}{n+2} =-\frac{1}{(n+2)(n+3)}<0. $$ Thus the sequence decreases, but never reaches $2$. In every tail, $2$ is a lower bound. Given any $\varepsilon>0$, choose an index $n\geq N$ large enough that $1/(n+2)<\varepsilon$. Then $a_n=2+1/(n+2)<2+\varepsilon$. No number greater than $2$ can therefore be a lower bound for the tail, so its infimum is $2$. Hence $i_N=2$ for every $N$, and $\liminf a_n=2$, although no term equals $2$.

The Two Infinite Cases

The infinite values of the liminf have different interpretations. A liminf of $-\infty$ means that every tail is unbounded below: arbitrarily negative terms continue to occur however far out we start. A liminf of $+\infty$, by contrast, means that the sequence tends to $+\infty$: every sufficiently late term eventually exceeds any specified real threshold.

Theorem (Infinite Liminf Criteria): For a real sequence $(a_n)$:

$\liminf_{n\to\infty}a_n=-\infty$ if and only if every tail $\{a_n:n\geq N\}$ is unbounded below.
$\liminf_{n\to\infty}a_n=+\infty$ if and only if $a_n\to+\infty$.

Proof. If $\sup_N i_N=-\infty$, then every $i_N\leq\sup_N i_N=-\infty$, since the supremum is an upper bound. Also $i_N\geq-\infty$, so every $i_N=-\infty$. Thus every tail is unbounded below. Conversely, if every tail is unbounded below, then every $i_N=-\infty$, and their supremum is $-\infty$.

Suppose next that $\sup_N i_N=+\infty$. Given any real $A$, there must be an $N$ with $i_N>A$; otherwise every $i_N\leq A$, so their supremum would be at most $A$. For every $n\geq N$, it follows that $$ a_n\geq i_N>A. $$ This is exactly $a_n\to+\infty$.

Conversely, suppose $a_n\to+\infty$. Given any real $A$, there is an $N$ such that $a_n>A$ for every $n\geq N$. Thus $A$ is a lower bound for that tail, and $i_N\geq A$. It follows that $\sup_j i_j\geq A$. Since this holds for every real $A$, $\sup_j i_j=+\infty$. $\square$

Worked Example: Liminf Equal to Negative Infinity Without Convergence to Negative Infinity

Define $a_n=-n$ when $n$ is even and $a_n=0$ when $n$ is odd. Every tail contains even indices as large as desired. Given any real lower bound $B$, choose an even $n\geq N$ with $n>-B$. Then $a_n=-n<B$. Thus no real number is a lower bound for any tail, so every tail infimum is $-\infty$ and $\liminf a_n=-\infty$.

This sequence does not tend to $-\infty$, because every tail also contains odd indices with value $0$. In particular, its terms are not eventually less than $-1$. The liminf condition here records repeated arbitrarily low values, not a claim that all sufficiently late terms are low.

Worked Example: Liminf Equal to Positive Infinity

Let $a_n=\sqrt{n+1}-2$. Given any real $A$, choose $N$ large enough that $\sqrt{N+1}>A+2$. The sequence is increasing, so for every $n\geq N$, $$ a_n=\sqrt{n+1}-2\geq\sqrt{N+1}-2>A. $$ Therefore $a_n\to+\infty$, and the Infinite Liminf Criteria give $\liminf a_n=+\infty$. In this case every sufficiently late term is above the threshold, which is stronger than merely having arbitrarily large terms in every tail.

Interpreting the Liminf Carefully

For a finite liminf $L$, the sequence is eventually above $L-\varepsilon$ for every positive $\varepsilon$, but every tail still contains a term below $L+\varepsilon$. These statements describe different aspects of the lower-tail behavior: one is an eventual bound on all terms, while the other guarantees that terms continue to approach the threshold from above or pass below it.

Do not confuse liminf equal to $-\infty$ with convergence to $-\infty$. The former requires every tail to be unbounded below; the latter requires all sufficiently late terms to lie below each real threshold. Similarly, a finite liminf need not be attained by any term, and it does not imply that the original sequence converges. The subsequence theorem gives a precise conclusion that does hold: a subsequence approaches every finite liminf.

Check Your Understanding

Use the tail-infimum definition and the proved criteria to answer the following questions.

Why are the tail infima $i_N$ nondecreasing as $N$ increases?
If a sequence has finite liminf $L$, what eventual bound holds relative to $L-\varepsilon$, and what must occur in every tail relative to $L+\varepsilon$?
How can a sequence have liminf $2$ even though no term equals $2$? Explain using $a_n=2+1/(n+2)$.
What condition on every tail characterizes liminf equal to $-\infty$?
Why does liminf equal to $+\infty$ imply that the sequence tends to $+\infty$?

What You'll Learn