Properties of liminf

From Tail Infima to Liminf Properties

The liminf describes the long-term lower behavior of a sequence through the infima of its tails. In the previous tutorial, corresponding properties of limsup were obtained from tail suprema. Here we use tail infima to study lower bounds, scalar multiplication, convergent perturbations, and sums. The direction of the inequalities matters: lower behavior reverses some of the familiar upper-behavior statements.

Recall from “The liminf of a Sequence” that the tail infimum of $(a_n)$ is

i_N(a)=\inf\{a_n:n\geq N\},

where an unbounded-below tail has infimum $-\infty$, and

\liminf_{n\to\infty}a_n=\sup_{N\in\mathbb{N}_0}i_N(a).

As $N$ increases, terms are removed from the tail, so its infimum cannot decrease: $i_{N+1}(a)\geq i_N(a)$. Thus the liminf is the supremum of an increasing sequence of tail infima, with the value allowed to be an extended real number. As with limsup, ignoring any fixed finite collection of initial tails does not change the result.

Eventual Comparison

If one sequence is eventually no smaller than another, then the infimum of each sufficiently late tail of the first sequence is at least the corresponding tail infimum of the second. Taking suprema over those tails preserves the inequality.

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

Proof. For each $N\geq N_0$, every term $a_n$ in the $N$-th tail is at least the corresponding term $b_n$. Therefore the infimum of the $N$-th tail of $(a_n)$ is at least that of $(b_n)$: $$ i_N(a)\geq i_N(b). $$ Taking suprema over $N\geq N_0$ gives the same inequality for the two liminfs. Restricting the supremum to $N\geq N_0$ does not change either liminf, since the tail infima are nondecreasing. Hence $$ \liminf_{n\to\infty}a_n\geq\liminf_{n\to\infty}b_n, $$ as claimed. $\square$

Worked Example: An Early Exception Does Not Affect the Comparison

Let $a_0=20$, and let $a_n=1+1/(n+1)$ for $n\geq1$. Let $b_n=2$ for every $n$. Although $a_0\geq b_0$, the eventual comparison we need is in the other direction: for $n\geq1$, $$ a_n=1+\frac{1}{n+1}\leq 2=b_n, $$ because $1/(n+1)\leq1$. The sequence $a_n$ decreases to $1$ for $n\geq1$, so the infimum of every tail starting at $N\geq1$ is $1$. Thus these tail infima are constantly $1$, and their limit is $1$. The tail infima of $(b_n)$ are all $2$. Thus $$ \liminf a_n=1\leq2=\liminf b_n. $$ The initial term $a_0$ does not affect the liminf or the eventual comparison.

Multiplication by a Scalar

Multiplication by a positive scalar preserves the order of terms, so it scales each tail infimum by that scalar. A negative scalar reverses the order: the infimum of the multiplied tail is the scalar times the supremum of the original tail. Consequently, negative multiplication connects liminf to limsup.

Theorem (Scalar Multiplication and Liminf): Let $c\in\mathbb{R}$. If $c>0$, then $$ \liminf_{n\to\infty}(ca_n)=c\liminf_{n\to\infty}a_n. $$ If $c<0$, then $$ \liminf_{n\to\infty}(ca_n)=c\limsup_{n\to\infty}a_n. $$ If $c=0$, then $\liminf_{n\to\infty}(ca_n)=0$. Products with extended values are interpreted with their usual order behavior.

Proof. Let $i_N(a)$ and $s_N(a)$ be the infimum and supremum of the $N$-th tail of $(a_n)$. If $c>0$, multiplication preserves order, so the infimum of the multiplied tail is $c i_N(a)$. Therefore $$ \liminf_{n\to\infty}(ca_n) =\sup_N\bigl(c i_N(a)\bigr) =c\sup_N i_N(a) =c\liminf_{n\to\infty}a_n. $$ If $c<0$, multiplication reverses order, so the infimum of the multiplied tail is $c s_N(a)$. Since multiplying by $c<0$ reverses order when taking the supremum over $N$, $$ \liminf_{n\to\infty}(ca_n) =\sup_N\bigl(c s_N(a)\bigr) =c\inf_N s_N(a) =c\limsup_{n\to\infty}a_n. $$ These identities also hold for unbounded tails using the corresponding extended values. Finally, if $c=0$, every term $ca_n$ is zero, so the liminf is zero. $\square$

Worked Example: A Negative Scalar Converts an Upper Limit to a Lower Limit

Define $a_n=1+(-1)^n$. For even $n$, $a_n=1+1=2$; for odd $n$, $a_n=1-1=0$. Every tail contains both parities, so its infimum is $0$ and its supremum is $2$. Hence $$ \liminf a_n=0,\qquad \limsup a_n=2. $$ For $c=-3$, the even-indexed terms of $(ca_n)$ equal $(-3)(2)=-6$, and the odd-indexed terms equal $(-3)(0)=0$. Every tail of the multiplied sequence has infimum $-6$, so $$ \liminf(-3a_n)=-6=(-3)\limsup a_n. $$ Using $\liminf a_n$ on the right instead would give $0$, which is not the liminf of the multiplied sequence.

Adding a Convergent Sequence

If $b_n\to L$, then sufficiently late terms of $(b_n)$ differ from $L$ by less than any prescribed positive error. Adding $b_n$ to $a_n$ therefore shifts the liminf by $L$, just as a constant shift does. The result applies whether the liminf of $(a_n)$ is finite or infinite.

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

Proof. For any fixed real number $c$, the infimum of the $N$-th tail of $(a_n+c)$ is $i_N(a)+c$. Taking suprema gives $$ \liminf_{n\to\infty}(a_n+c)=I+c. $$ Now fix $\varepsilon>0$. Since $b_n\to L$, there is an $N_0$ such that $|b_n-L|<\varepsilon$ for every $n\geq N_0$. Thus, for every such $n$, $$ a_n+L-\varepsilon<a_n+b_n<a_n+L+\varepsilon. $$ By Order Preservation for Liminf, and by the constant-shift identity just established, $$ I+L-\varepsilon \leq\liminf_{n\to\infty}(a_n+b_n) \leq I+L+\varepsilon. $$ If $I$ is finite, these bounds for every $\varepsilon>0$ force the middle value to equal $I+L$. If $I=+\infty$, the lower bound says the middle liminf is at least $+\infty$, and so it equals $+\infty$. If $I=-\infty$, the upper bound says the middle liminf is at most $-\infty$, and so it equals $-\infty$. This proves the identity in all cases. $\square$

Worked Example: A Convergent Perturbation Shifts the Liminf

Let $a_n=7+(-1)^n$ and $b_n=3+1/(n+1)$. The sequence $a_n$ equals $8$ at even indices and $6$ at odd indices, so every tail has infimum $6$, giving $\liminf a_n=6$. Also, $b_n\to3$. The perturbation theorem therefore gives $$ \liminf(a_n+b_n)=6+3=9. $$ To check this directly, for even $n$ the sum is $11+1/(n+1)$, and for odd $n$ it is $9+1/(n+1)$. Every tail contains odd indices as large as desired, at which these values approach $9$ from above. The even-indexed values are all greater than $11$, so they do not lower the tail infimum. The tail infima consequently approach $9$, in agreement with the theorem.

A Lower Bound for the Liminf of a Sum

If both sequences have finite liminfs, then their sum has liminf at least the sum of those two values. Indeed, sufficiently late terms of each sequence lie above its liminf minus a small error, so sufficiently late sums lie above the sum of the liminfs minus the combined error. Equality can fail when the sequences do not approach their lower behavior at the same indices.

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

Proof. Fix $\varepsilon>0$. By the Threshold Characterization of a Finite Liminf, 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 the two inequalities gives $$ a_n+b_n>A+B-2\varepsilon. $$ Order Preservation for Liminf, applied to this eventual comparison with the constant sequence $A+B-2\varepsilon$, yields $$ \liminf_{n\to\infty}(a_n+b_n)\geq A+B-2\varepsilon. $$ This holds for every $\varepsilon>0$. If the liminf on the left were less than $A+B$, choose $\varepsilon$ so small that $2\varepsilon$ is less than the difference between $A+B$ and that liminf. The displayed inequality would then be contradicted. Therefore the liminf is at least $A+B$. $\square$

Worked Example: The Liminf Inequality Can Be Strict

Set $a_n=(-1)^n$ and $b_n=-(-1)^n$. Each sequence takes the values $1$ and $-1$ infinitely often. Every tail of each therefore has infimum $-1$, and $$ \liminf a_n=-1,\qquad \liminf b_n=-1. $$ But $a_n+b_n=(-1)^n-(-1)^n=0$ for every $n$, so $$ \liminf(a_n+b_n)=0>-2=\liminf a_n+\liminf b_n. $$ At each index, one sequence equals $1$ while the other equals $-1$. Their lower values never occur together, which is why the sum stays above the sum of the individual liminfs.

Using Liminf Properties Carefully

The direction of comparison is a useful check: if $a_n\geq b_n$ eventually, then the liminf of $(a_n)$ is at least the liminf of $(b_n)$. Positive multiplication preserves this lower behavior, whereas negative multiplication reverses it and requires the limsup. A convergent perturbation shifts the liminf by the perturbation's limit, and the sum inequality supplies a lower bound rather than an automatic equality.

A common error is to infer that the liminf of a sum must equal the sum of the liminfs. The strict example shows that the two sequences may attain their lower values at different indices. Another is to apply the finite-liminf sum inequality when one of the liminfs is infinite without checking the extended-real expression: combinations such as $+\infty+(-\infty)$ are not defined. Verify the hypotheses before using the inequality.

Check Your Understanding

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

If $a_n\geq b_n$ for all sufficiently large $n$, what comparison follows for their liminfs?
For $c<0$, which limiting quantity determines $\liminf(ca_n)$, and why does the order reverse?
If $b_n\to L\in\mathbb{R}$, how is $\liminf(a_n+b_n)$ related to $\liminf a_n$, including when the latter is infinite?
State the liminf inequality for a sum when both individual liminfs are finite.
Why can the liminf of a sum be strictly greater than the sum of the individual liminfs?

What You'll Learn