limsup, liminf and subsequential limits

In one sentence

Every bounded sequence has a highest and a lowest place its terms keep returning to, and those two numbers exist even when the limit does not.

Why the exam asks it

Because limsup is the tool that lets you say something definite about a sequence that refuses to converge — and the paper likes sequences that refuse to converge. It also has an algebra with one inequality that is strict often enough to be a reliable distractor, which is worth more to an examiner than any computation.

The idea

Take a bounded sequence $(a_n)$ and look at its tail from position $n$ onwards. That tail has a supremum, $s_n = \sup_{k \ge n} a_k$.

Now push $n$ forward. The tail gets shorter, so its supremum cannot increase — you are taking the sup of a smaller set. So $(s_n)$ is decreasing, and being bounded below it converges. That limit is the limsup:

$$\limsup_{n} a_n = \lim_{n \to \infty} \sup_{k \ge n} a_k = \inf_n \sup_{k \ge n} a_k.$$

The liminf is the same construction with infima, rising instead of falling.

There are two other ways to say the same thing, and you want all three.

As subsequential limits. The limsup is the largest limit any subsequence achieves; the liminf is the smallest. For $a_n = (-1)^n$ the subsequential limits are $1$ and $-1$, so $\limsup = 1$ and $\liminf = -1$.

As an $\varepsilon$-statement. $L = \limsup a_n$ means: for every $\varepsilon > 0$,

• $a_n < L + \varepsilon$ eventually — from some point onward, all terms are below;
• $a_n > L - \varepsilon$ infinitely often — the sequence keeps coming back up near $L$.

This third form is what most Part C statements are actually testing. Note the asymmetry: one "eventually", one "infinitely often". Options that swap those two quantifiers are wrong, and they are wrong in a way that reads perfectly well.

The algebra

$$\liminf a_n \le \limsup a_n, \quad \text{with equality (and finite) exactly when } a_n \text{ converges}.$$

That is the useful characterisation of convergence for these questions. Also:

• $\limsup(-a_n) = -\liminf a_n$.
• $\limsup(a_n + b_n) \le \limsup a_n + \limsup b_n$ — an inequality, not an identity.
• If $b_n \to b$ converges, then $\limsup(a_n + b_n) = \limsup a_n + b$. Equality returns once one of the two settles down.
• $\limsup a_n = +\infty$ exactly when $(a_n)$ is unbounded above.

Where intuition breaks

"$\limsup(a_n + b_n) = \limsup a_n + \limsup b_n$." The single most productive false statement in this topic. Take $a_n = (-1)^n$ and $b_n = (-1)^{n+1}$. Their sum is identically zero, so the left side is $0$. But each has limsup $1$, so the right side is $2$.

Why it is tempting: limits are additive, and limsup looks like a limit. It is not — it is a sup, and sups only ever satisfy an inequality, because the two sequences can be arranged never to peak together. That is the whole content of the failure.

"limsup is the supremum of the sequence." No — it ignores every finite chunk. For $a_n = 1/n$ the supremum is $1$ and the limsup is $0$. The limsup cares only about the tail, which is exactly why it is stable under changing finitely many terms.

"limsup = liminf, so the sequence converges." True when both are finite. If both are $+\infty$ they are equal and the sequence diverges to infinity. Options omitting the finiteness are testing precision, not understanding, and they cost the same mark either way.

"The sequence gets arbitrarily close to $L$, so $\limsup = L$." Getting close infinitely often makes $L$ a subsequential limit — so $L \le \limsup$. It does not make $L$ the largest one. $a_n = (-1)^n$ comes arbitrarily close to $-1$ infinitely often, and its limsup is $1$.

"$\sin n$ oscillates between $-1$ and $1$ but never settles, so its limsup is undefined." It is exactly $1$. The set $\{n \bmod 2\pi\}$ is dense in $[0, 2\pi)$, so $\sin n$ comes arbitrarily close to every value in $[-1,1]$, and the largest subsequential limit is $1$. Limsup is defined for every bounded sequence, however badly behaved — that is the point of it.

The exam's angle

1. Given an explicit sequence, compute both. Split it into its natural tracks — even and odd terms, or by residue class — take the limit along each, and read off the largest and smallest.
2. Given a statement about the algebra, hunt for strictness. If an option asserts equality where the theorem gives $\le$, try $(-1)^n$ against $(-1)^{n+1}$ before anything else.
3. Given "the sequence converges if and only if …", check that finiteness is stated.

For step 1, the two-track method handles most of what appears: sequences built from $(-1)^n$, from $n \bmod k$, or from a slowly varying factor times an oscillating one.

The night before

• $\limsup = \lim_n \sup_{k \ge n} a_k$. The tail sup falls; the tail inf rises.
• Equivalently: the largest subsequential limit.
• $\varepsilon$-form: below $L + \varepsilon$ eventually, above $L - \varepsilon$ infinitely often. The quantifiers differ, and that difference is examinable.
• Converges $\iff$ limsup $=$ liminf and finite.
• Subadditive only: $\limsup(a_n + b_n) \le \limsup a_n + \limsup b_n$. Spoiler is $(-1)^n$ against $(-1)^{n+1}$, giving $0 < 2$.
• Equality returns if one sequence converges.
• $\limsup \sin n = 1$, because $\{n \bmod 2\pi\}$ is dense.