The Definitions
Limit Superior
$\limsup s_n = \lim_{N \to \infty} \sup \{s_n : n \gt N\}$
Limit Inferior
$\liminf s_n = \lim_{N \to \infty} \inf \{s_n : n \gt N\}$
Intuition: Imagine cutting off the first $N$ terms of the sequence. What is the supremum (least upper bound) of what's left? As you cut off more and more terms ($N \to \infty$), that supremum settles down to a value. That value is the $\limsup$.
Visualizing the "Tail"
Drag slider to increase N (ignore first N terms)
Key Properties & Theorems
Subsequential Limits
$\limsup s_n$ is the largest subsequential limit.
$\liminf s_n$ is the smallest subsequential limit.
Convergence Test
A sequence $(s_n)$ converges if and only if: $$\liminf s_n = \limsup s_n$$
Always Exists
Unlike the regular limit $\lim s_n$, the $\limsup$ and $\liminf$ always exist (possibly $\pm \infty$) for any sequence of real numbers.
Example Walkthrough
Consider the sequence $s_n = (-1)^n + \frac{1}{n}$. Let's find the lim sup and lim inf.
Step 1: Write out terms
- n=1: -1 + 1 = 0
- n=2: 1 + 1/2 = 1.5
- n=3: -1 + 1/3 = -0.66...
- n=4: 1 + 1/4 = 1.25
- n=5: -1 + 1/5 = -0.8
Step 2: Identify Subsequences
The even terms are $1 + 1/2k \to 1$.
The odd terms are $-1 + 1/(2k-1) \to -1$.
Conclusion:
The set of subsequential limits is $S = \{-1, 1\}$.
$\limsup s_n = \sup S = 1$
$\liminf s_n = \inf S = -1$
Practice Problems
True/False
1. $\limsup s_n$ always exists (possibly $\pm\infty$) for any sequence of real numbers.
True. Every set of real numbers has a supremum in the extended real number system, and the limit of a monotone sequence always exists in the extended reals.
2. If $\limsup s_n = \liminf s_n$, then the sequence converges to a finite limit.
False. The sequence could diverge to $+\infty$ or $-\infty$. If they are equal and finite, then it converges.
3. $\limsup (s_n + t_n) \le \limsup s_n + \limsup t_n$ (assuming the sum is defined).
True. This is a standard property of limit superior.
4. If a sequence is bounded, then $\limsup s_n$ is a real number.
True. Boundedness implies the set of subsequential limits is bounded, so its supremum is finite.
Fill in the Blank
1. The limit superior of a sequence $(s_n)$ is the limit of the sequence $v_N = \sup \{s_n : n \gt N\}$. The sequence $v_N$ is always ______.
Monotone decreasing (or non-increasing). As $N$ increases, the set $\{s_n : n \gt N\}$ becomes smaller (subset of previous), so its supremum cannot increase.
2. If $S$ is the set of subsequential limits of $(s_n)$, then $\limsup s_n = \sup S$. If $(s_n)$ is bounded, then $\limsup s_n$ is the ______ element of $S$.
Largest (or maximum). The set of subsequential limits is closed, so it contains its supremum.
3. A sequence $(s_n)$ converges to $L \in \mathbb{R}$ if and only if $\liminf s_n = \limsup s_n = $ ______.
L.
Full Problems
1. Find $\limsup s_n$ and $\liminf s_n$ for $s_n = (-1)^n + \frac{1}{n}$.
The terms are roughly $1, -1, 1, -1, \dots$. Specifically, for even $n$, $s_n \to 1$. For odd $n$, $s_n \to -1$. Thus, the set of subsequential limits is $\{1, -1\}$. So $\limsup s_n = 1$ and $\liminf s_n = -1$.
2. Let $s_n = \sin(n\pi/3)$. Find $\limsup s_n$.
The sequence of values repeats: $\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}, 0, -\frac{\sqrt{3}}{2}, -\frac{\sqrt{3}}{2}, 0, \dots$. The set of subsequential limits is $\{-\frac{\sqrt{3}}{2}, 0, \frac{\sqrt{3}}{2}\}$. Thus, $\limsup s_n = \frac{\sqrt{3}}{2}$.
3. Give an example of a sequence where $\liminf s_n \lt \limsup s_n$.
Any oscillating sequence works. For example, $s_n = (-1)^n$. $\liminf s_n = -1$ and $\limsup s_n = 1$.
4. Prove that if $s_n \le t_n$ for all $n$, then $\limsup s_n \le \limsup t_n$.
Let $u_N = \sup \{s_n : n \gt N\}$ and $v_N = \sup \{t_n : n \gt N\}$. Since $s_n \le t_n$, we have $\{s_n : n \gt N\} \subseteq (-\infty, v_N]$, so $u_N \le v_N$. Taking the limit as $N \to \infty$, we get $\lim u_N \le \lim v_N$, which means $\limsup s_n \le \limsup t_n$.