2.3 Limit superior and limit inferior

Look at the tails

For a bounded real sequence $(x_n)$, define

$$u_N=\sup\{x_n:n\ge N\},\qquad \ell_N=\inf\{x_n:n\ge N\}.$$

The tail sets shrink as $N$ increases, so $(u_N)$ is decreasing and $(\ell_N)$ is increasing. Both are bounded, and therefore converge. Define

$$\limsup_{n\to\infty}x_n=\lim_{N\to\infty}u_N=\inf_N u_N, \qquad \liminf_{n\to\infty}x_n=\lim_{N\to\infty}\ell_N=\sup_N\ell_N.$$

A finite set of early terms can change the first few tail bounds but not either limit. Extended values can be used for unbounded sequences, but this unit restricts to bounded sequences so every displayed quantity is real.

Worked example

Oscillation with a vanishing correction

Let $x_n=(-1)^n+1/n$. The even subsequence tends to $1$ and the odd subsequence to $-1$. Since every $x_n\le1+1/n$, the tail supremum is at most $1+1/N$; even terms arbitrarily far out approach $1$, so it is at least $1$. Thus $u_N\to1$. Similarly $-1\le\ell_N\le-1+1/n$ for arbitrarily large odd $n$, giving $\ell_N\to-1$. Therefore $\limsup x_n=1$ and $\liminf x_n=-1$.

Theorem

Subsequential characterization

For a bounded real sequence, the limit superior is the largest subsequential limit and the limit inferior is the smallest subsequential limit. The whole sequence converges if and only if its limit superior and limit inferior are equal; then their common value is the limit.

Proof

Construct subsequences and squeeze the whole sequence

Let $U=\lim u_N$ and set $n_0=0$. Choose increasing indices $n_k\ge k$ with $x_{n_k}>u_{n_{k-1}+1}-1/k$; this is possible by the supremum property of the relevant tail. Since its starting index tends to infinity, its supremum tends to $U$, while $x_{n_k}\le u_k$ after choosing $n_k\ge k$. Thus $x_{n_k}\to U$. If any subsequence tends to $L$, then for each fixed $N$ all sufficiently late subsequence terms lie in the $N$th tail, so $L\le u_N$. Hence $L\le U$. Applying the argument to $-x_n$ gives the corresponding infimum conclusion.

For every $n\ge N$, $\ell_N\le x_n\le u_N$. If both bounds tend to the same value, the squeeze principle proves $x_n$ converges to it. Conversely, convergence of the original sequence forces every subsequential limit to equal its limit, so the limsup and liminf agree.

Worked example

A convergent sequence with one large early term

Set $x_1=100$ and $x_n=1/n$ for $n\ge2$. The supremum of the entire set of terms is $100$, but $\limsup x_n=\liminf x_n=0$. Limit superior concerns arbitrarily late behavior rather than the largest term ever observed.

Method comparison

For explicit alternating patterns, compute limits of their component subsequences and bound all terms. For a direct definition proof, work with tail suprema and infima. To prove ordinary convergence, equalize limsup and liminf; to disprove it, exhibit distinct subsequential limits. Do not equate a global supremum with a tail limit.

Practice

1. Find limsup and liminf of $(-1)^n$.
2. Find limsup and liminf of $1/n$.
3. Find limsup and liminf of $x_n=2+(-1)^n/n$.
4. Explain why a single large initial term does not change limsup.

Full solutions

1. Every tail contains both $1$ and $-1$. Its supremum is $1$ and infimum is $-1$, so limsup is $1$ and liminf is $-1$.
2. Every tail has supremum $1/N$ and infimum $0$, so both limits are $0$.
3. Since $|x_n-2|\le1/n\to0$, $x_n\to2$. Both limsup and liminf equal $2$.
4. Deleting or changing finitely many terms leaves every sufficiently late tail unchanged. Thus the limits of tail suprema and infima are unchanged.

Synthesis

Limsup and liminf capture the largest and smallest limits accessible along subsequences. They coincide precisely when the whole bounded sequence converges.