limsup, liminf, and Subsequential Limits: Examples and Counterexamples

Let $(s_k)$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$, the sequence usually written $s_k = (-1)^k$, characterised by $s_0 = 1$ and $s_{k+1} = -s_k$. Then

$\liminf_{k} s_k = -1, \qquad \limsup_{k} s_k = 1,$

so the two differ and $(s_k)$ neither converges nor diverges to $\pm\infty$ (A real sequence converges to $L \in \mathbb{R}$ iff $\liminf x_k = \limsup x_k = L$, and diverges to $\pm\infty$ iff both equal $\pm\infty$).

This is the smallest example in which the inequality $\liminf \le \limsup$ is strict, and it shows exactly what the gap measures: the sequence keeps returning to two different values, and neither of them can be the limit because the other keeps interrupting.

• The two values are exactly the subsequential limits. By The limit superior is itself a subsequential limit in $\overline{\mathbb{R}}$ and is the greatest one and The limit inferior is the least subsequential limit in $\overline{\mathbb{R}}$ the numbers $1$ and $-1$ are the greatest and least elements of $\overline{\operatorname{SL}}(s)$, and since every term is $1$ or $-1$ no other value can be a subsequential limit, so $\operatorname{SL}(s) = \{-1, 1\}$ exactly.

• Contrast with The sequence $(-1)^k(1 + 1/k)$ is bounded with subsequential limit set exactly $\{-1, 1\}$. There the same two subsequential limits arise for a sequence none of whose terms equals either of them. The limit superior and limit inferior do not care: they are determined by the tails, not by whether the values are attained.

• This sequence is the standard witness for strictness throughout the page. It drives FALSE: $\limsup(x_k + y_k) = \limsup x_k + \limsup y_k$ and $x_k = (-1)^k$, $y_k = (-1)^{k+1}$ give $\limsup(x_k + y_k) = 0 < 2 = \limsup x_k + \limsup y_k$, and, after an affine change, $x_k = 1 + (-1)^k$, $y_k = 1 + (-1)^{k+1}$ give $\limsup(x_k y_k) = 0 < 4$.

Write $\iota(n) = n \cdot 1_{\mathbb{R}}$ for the canonical natural. Let $c : \mathbb{N} \to \mathbb{N} \times \mathbb{N}$ be the function supplied by the recursion theorem (The recursion theorem) from the starting element $(1,1)$ and the function

$F(m, j) := (m,\, j+1) \ \text{ if } j < m, \qquad F(m, j) := (m+1,\, 1) \ \text{ if } j = m,$

write $c_k = (m_k, j_k)$, and define the sequence of reals

$x_k \;:=\; \frac{\iota(j_k)}{\iota(m_k)} \qquad (k \in \mathbb{N}).$

Its first terms are

$\frac{1}{1};\quad \frac{1}{2}, \frac{2}{2};\quad \frac{1}{3}, \frac{2}{3}, \frac{3}{3};\quad \frac{1}{4}, \dots$

the blocks being the successive values of $m$. Then the subsequential limit set (Subsequential limit of a real sequence, and the subsequential limit set) is the whole unit interval (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length),

$\operatorname{SL}(x) = [0,1],$

and consequently $\limsup_k x_k = 1$ and $\liminf_k x_k = 0$ (Limit superior and limit inferior of a real sequence as $\inf_n \sup_{k \ge n} x_k$ and $\sup_n \inf_{k \ge n} x_k$ in $\overline{\mathbb{R}}$).

The recursion replaces the block bookkeeping. Presenting the sequence by the partial sums $1 + 2 + \dots + (m-1)$ that mark where each block begins would require inverting that count at every index. Carrying the pair $(m, j)$ along instead makes each term's block and position immediately available, and the three facts the argument needs, that $1 \le j_k \le m_k$, that $m_k \le k+1$, and that every admissible pair occurs, are then three short inductions.

• Every point of $[0,1]$ is approached, and the rate is the block width. In block $m$ the terms are $1/m, 2/m, \dots, m/m$, spaced $1/m$ apart and covering $(0,1]$, so any target in $[0,1]$ has a term of block $m$ within $1/m$ of it. Since blocks of every width occur, and occur arbitrarily late, this produces a subsequence converging to the target. That is the whole idea; steps 2.2 and 3.2 only make the choice of term canonical, by taking a least element rather than an arbitrary one, so that no choice principle is used.

• The set is closed, as it must be. $[0,1]$ contains the limit of every convergent sequence of its own points, which is what If each $y_j$ is a subsequential limit of $(x_k)$ and $y_j \to y \in \mathbb{R}$, then $y$ is a subsequential limit of $(x_k)$ predicts for any subsequential limit set. This example shows the prediction is not vacuous: the set here is an entire interval, in contrast with the two-point set of $(-1)^k$ has $\liminf = -1$ and $\limsup = 1$, so it does not converge.

• Neither endpoint is a value of the sequence in the case of $0$. Every term is $> 0$ by step 2.1, so $0 \in \operatorname{SL}(x)$ is a genuine limit and not an attained value, while $1$ is attained, once in every block. Subsequential limits need not be values, and values need not be subsequential limits.

• Why the sequence is not written by a closed formula. The classical presentation defines $x_k$ by first solving $m(m-1)/2 \le k < m(m+1)/2$ for $m$, which needs a least-element argument at every index and a quadratic estimate to get $m \le k+1$. The recursion carries the block and position forward instead, and the estimate $m_k \le k+1$ becomes the one-line induction of step 1.3.

Let $(s_k)$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$, let $t_k := 2$ when $s_k = 1$ and $t_k := 1/2$ when $s_k = -1$, and put

$a_k \;:=\; 2^{-k} t_k \qquad (k \in \mathbb{N}),$

the sequence usually written $a_k = 2^{-k + (-1)^k}$. Writing $q_k := a_{k+1}/a_k$ for the ratios and $r_k := a_{k+1}^{1/(k+1)}$ for the roots, which is $a_n^{1/n}$ reindexed by $n = k+1$ as For $a_k > 0$: $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$ requires,

$\liminf_{k} q_k = \frac{1}{8}, \qquad \limsup_{k} q_k = 2, \qquad \lim_{k} r_k = \frac{1}{2},$

so also $\liminf_k r_k = \limsup_k r_k = 1/2$. In addition $a_k \to 0$.

The point. The ratios oscillate across $1$, taking the values $1/8$ and $2$ alternately, so no statement of the form "the ratios are eventually below some $\lambda < 1$" is available; the roots, by contrast, converge to $1/2 < 1$. Any criterion reading the ratios alone is silent here, and one reading the roots is not. That is the concrete form of the dominance recorded in For $a_k > 0$: $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$.

• Where the numbers come from. The exponent $-k + (-1)^k$ changes by $-1 + (-1)^{k+1} - (-1)^k = -1 \mp 2$ from one index to the next, giving ratios $2^{-3} = 1/8$ and $2^{1} = 2$; the root divides the exponent by the index, so the bounded oscillation contributes $2^{\pm 1/(k+1)} \to 1$ and only the linear part $-k$ survives, giving $2^{-1} = 1/2$. The full computation is in FALSE: $\limsup a_k^{1/k} = \limsup a_{k+1}/a_k$ for every positive sequence.

• The same sequence reappears for series. With these $a_k$ the series $\sum_k a_k$ converges, and the root criterion sees it while the ratio criterion does not. That use belongs to the series page and is not made here.

• Strictness of the middle inequality needs a different witness. Here $\liminf_k r_k = \limsup_k r_k$, since the roots converge. A sequence making all three inequalities of the chain strict is A positive sequence making all three inequalities of the ratio-to-root chain strict.

Let $(s_k)$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$, $s_{\sigma(k)} = -s_k$, which satisfies $|s_k| = 1$, $s \circ e \equiv 1$ and $s \circ o \equiv -1$ and define

$a_k := 2^{-k} \ \text{ when } s_k = 1, \qquad a_k := 3^{-k} \ \text{ when } s_k = -1 .$

This interleaves the two geometric sequences $2^{-k}$ and $3^{-k}$, taking the first at even indices and the second at odd ones. With $q_k := a_{k+1}/a_k$ and $r_k := a_{k+1}^{1/(k+1)}$ as in For $a_k > 0$: $\liminf a_{k+1}/a_k \le \liminf a_k^{1/k} \le \limsup a_k^{1/k} \le \limsup a_{k+1}/a_k$,

$\liminf_{k} q_k = 0, \qquad \liminf_{k} r_k = \frac{1}{3}, \qquad \limsup_{k} r_k = \frac{1}{2}, \qquad \limsup_{k} q_k = +\infty,$

so the chain of that theorem reads

$0 \;<\; \tfrac{1}{3} \;<\; \tfrac{1}{2} \;<\; +\infty$

with all three inequalities strict.

Where each comparison lives. The first two, $0 < 1/3$ and $1/3 < 1/2$, are comparisons of real numbers and hold in $\mathbb{R}$; they hold in $\overline{\mathbb{R}}$ as well only because the extended order restricts on $\mathbb{R}$ to the order of $\mathbb{R}$ (The extended real line $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, +\infty\}$, its order, and the arithmetic that is left undefined). The third, $1/2 < +\infty$, is not a comparison in $\mathbb{R}$ at all: $+\infty$ is not a real number, and the inequality is the instance of "every real is below the greatest element" in $\overline{\mathbb{R}}$. So the outer two values of the chain are of different kinds here, and only the extended line can hold all four at once.