A divergent bounded sequence still has a largest and a smallest place where its terms pile up. These two numbers, the limit superior and limit inferior, always exist for bounded sequences (by Bolzano–Weierstrass and completeness), and they will power the root test and the radius of convergence later.
Key idea
For bounded $(a_n)$, let $s_n = \sup\{a_k : k \ge n\}$ (the tail supremum) and $i_n = \inf\{a_k : k \ge n\}$. Then $\limsup a_n = \lim s_n$ and $\liminf a_n = \lim i_n$. They are the largest and the smallest subsequential limits, and $a_n$ converges exactly when $\limsup a_n = \liminf a_n$ (then both equal the limit).
Why it works
As $n$ grows the tail $\{a_k : k \ge n\}$ loses elements, so its supremum can only go down: $(s_n)$ is decreasing, and it is bounded below. By monotone convergence it has a limit. The same works for $(i_n)$, which increases. Dropping finitely many terms never changes these limits, which is why they describe the long-run behavior and ignore early outliers. If $\limsup = \liminf = L$, then $i_n \le a_n \le s_n$ and both bounds tend to $L$, so $a_n \to L$ by the squeeze theorem.
Worked examples
Worked example: Tails of an alternating sequence
For $a_n = (-1)^{n}\left(1 + \frac{2}{n}\right)$, find $\limsup a_n$, $\liminf a_n$, $\sup a_n$ and $s_5$.
- Even $n$: $1 + \frac{2}{n} \to 1$ from above. Odd $n$: $-1 - \frac{2}{n} \to -1$ from below.
- So $\limsup a_n = 1$ and $\liminf a_n = -1$.
- The largest term overall is $a_2 = 2$, so $\sup a_n = 2$. From $k = 5$ on, the largest term is at the first even index, $k = 6$: $s_5 = 1 + \frac{2}{6} = \frac{4}{3}$.
Answer: $\limsup = 1$, $\liminf = -1$, $\sup = 2$, $s_5 = \frac{4}{3}$.
Worked example: A sequence that cycles
Find $\limsup$ and $\liminf$ of $a_n = (n \bmod 3) + \frac{1}{n}$.
- The remainders cycle $1, 2, 0, 1, 2, 0, \ldots$, and $\frac{1}{n} \to 0$.
- The subsequential limits are $0$, $1$ and $2$.
Answer: $\limsup = 2$, $\liminf = 0$.
For an unbounded sequence one writes $\limsup a_n = \infty$ when the terms are not bounded above. The examples in this course are bounded, so both numbers are finite.
Common mistake
Confusing $\sup$ with $\limsup$. The supremum of all the terms can be an early term that is never approached again, like $a_2 = 2$ above. The lim sup forgets any finite number of terms: it is where the tail suprema end up. Check by asking: do infinitely many terms come close to this value?