# Sequence converges iff $\limsup = \liminf$

I want to prove that a sequence of real numbers $\{s_n\}$ converges to $s$ if and only if $\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s$.

Here are my definitions:

For any sequence of real numbers $\{s_n\}$, let $E$ be the set of all subsequential limits of $\{s_n\}$, including possibly $+\infty$ and/or $-\infty$ if any subsequence of $\{s_n\}$ diverges to infinity. Then $\limsup_{n \to \infty} s_n = \sup E$, and $\liminf_{n \to \infty} s_n = \inf E$.

I know the theorem that a sequence converges to a point if and only if every one of its subsequences converge to that same point, so one direction of this proof is easy:

If $\{s_n\}$ converges to some point $s \in \mathbb{R}$, then every subsequence of $\{s_n\}$ converges to $s$. So the set $E$ of every subsequential limit of $\{s_n\}$ consists of the single point $s$, so $$\limsup_{n \to \infty} s_n = \sup \{s\} = s = \inf \{s\} = \liminf_{n \to \infty} s_n$$

But the other direction seems more tricky...

If $\limsup_{n \to \infty} s_n = \liminf_{n \to \infty} s_n = s$, then every convergent subsequence converges to the same point $s$. Also, there can be no subsequences which diverge to infinity (otherwise $\limsup_{n \to \infty} s_n$ would be $+\infty$, or $\liminf_{n \to \infty} s_n$ would be $-\infty$).

But can't there be subsequences which diverge otherwise? And wouldn't that throw off the convergence of $\{s_n\}$?

EDIT:

I'd also be willing to accept a solution which makes use of the "Pinching Theorem" (if $a_n \leq s_n \leq b_n$ for every $n \in \mathbb{N}$, and if $a_n \to s$ and $b_n \to s$, then $s_n \to s$).

• There are a few ways to reason this but one way can proceed by contradiction: suppose that $s_n$ does not converge to $s$, then there exists an $\epsilon>0$ such that... – Alex R. Mar 21 '12 at 1:56
• Wish I could upvote this more... oh well. Kudos for including the definitions, showing your work, using LaTex, etc. – The Chaz 2.0 Mar 21 '12 at 2:01

## 3 Answers

Useful facts that you should verify:

1. Any unbounded sequence has a subsequence diverging to $\infty$ or to $-\infty$.

2. Any bounded sequence has a convergent subsequence.

You correctly point out that the hypothesis that $\limsup_{n \to \infty} s_n$ and $\liminf_{n \to \infty} s_n$ are both finite implies that $(s_n)_{n=1}^{\infty}$ has no subsequences that diverge to infinity. But (1) implies that more is true: the sequence $(s_n)_{n=1}^{\infty}$ must be bounded.

Fix any $\epsilon > 0$.

• There cannot be infinitely many $n$ for which $s_n \geq s + \epsilon$, because you could select out of them a subsequence $y_k = s_{n_k}$ satisfying $y_k \geq s + \epsilon$ for all $k$.

• Since the sequence $(s_n)_{n=1}^{\infty}$ is bounded, so is the sequence $(y_k)_{k=1}^{\infty}$. So by (2) it has a subsequence convergent to some limit $L$; and by basic facts about limits, since $y_k \geq s + \epsilon$ for all $k$, one must have $L \geq s + \epsilon$.
• But $L$ is clearly also a subsequential limit of $s$, allowing us to deduce that $\limsup_{n \to \infty} s_n \geq s + \epsilon$, a contradiction.
• Similarly, there cannot be infinitely many $n$ for which $s_n \leq s - \epsilon$.

So there is a positive integer $N$ with the property that whenever $n \geq N$ one has $$s - \epsilon < s_n < s + \epsilon,$$ and since $\epsilon > 0$ was arbitrary, $(s_n)_{n=1}^{\infty}$ converges to $s$.

• Hi! I've read your proof and I have a question, does it change something if the series $s_n \in \mathbb{C}$? – Bman72 Mar 14 '14 at 16:01
• @Ale "Useful fact 2" of my answer holds true for complex sequences without change. It is unclear how to make sense of "diverging to $\infty$ or $-\infty$" so that "useful fact 1" remains both meaningful, useful, and true. And since $\mathbb{C}$ is not an ordered field, it's not immediately clear how to make sense of the original problem (as it's not clear how to define $\limsup$ and $\liminf$ for complex sequences --- and however you do try to define them, it may be difficult to prove that e.g. a bounded sequence of complex numbers has a subsequence converging to its $\limsup$...) – leslie townes Mar 15 '14 at 6:12
• @Ale But if $(s_n)$ is a complex sequence, and $(x_n)$ and $(y_n)$ are the real sequences defined by $s_n = x_n + i y_n$ for all $n$, then it's true that $(s_n)$ converges in $\mathbb{C}$ to a complex number $s = a + bi$ (with $a, b$ real) if and only if $\limsup x_n = \liminf x_n = a$ and $\limsup y_n = \liminf y_n = b$. (The key is the general fact that $(s_n)$ converges in $\mathbb{C}$ if and only if both $(x_n)$ and $(y_n)$ converge in $\mathbb{R}$: this should be easy to prove from whatever your definition of convergence in $\mathbb{C}$ is. Then apply the result of this problem, twice.) – leslie townes Mar 15 '14 at 6:14

Another possible solution (for bounded case only), just use these facts:

1. $\inf A=\sup A \quad\Leftrightarrow\quad A=\{\text{single element}\}.$ (A bounded and nonempty) -Proof

2. $(x_n)\text{ bounded and every convergent subsequence of } x_n\text{ converges to } a \quad\Rightarrow \quad\lim(x_n)=x$$-Proof Here's my proof based on more basic observations. Setup: For a sequence$(s_n), n \in \mathbb{N}$, if we define$s^+_n = \sup_{j\geq n}(s_j)$and$s^-_n = \inf_{j\geq n}(s_j)$, then we can equivalently define$\lim \sup_{n\to \infty} s_n := \inf_n s^+_n$and$\lim \inf_{n\to \infty} s_n := \sup_n s^-_n$(this is actually how my analysis book by Terence Tao defined them; also see this discussion). I'll use$L^+=\lim \sup_{n\to \infty} s_n$and$L^-=\lim \inf_{n\to \infty} s_n$for convenience. Now on to the proof: 1. If a sequence$(s_n)$converges to$s\in\mathbb{R}$, then$\lim \sup_{n\to\infty} s_n=\lim \inf_{n\to\infty} s_n =s$. Proof: Let's just prove$\lim \sup_{n\to\infty} s_n=s$; then$\lim \inf_{n\to\infty} s_n =s$can be proved similarly. Since$(s_n)$converges,$(s_n^+)$must also converge (because it's also bounded, and is also monotonically decreasing), then it's not hard to show$\lim \sup_{n\to\infty} s_n = \inf_n s^+_n = \lim_{n\to \infty} s^+_n = L^+ $. Note that$\forall n \in \mathbb{N}, s_n \leq s^+_n$, therefore$\lim_{n\to\infty} s_n \leq \lim_{n\to \infty} s^+_n $, i.e.,$s \leq L^+$. Now we just need to show it's not the case that$s < L^+$. For contradiction, suppose$s < L^+$. Since$(s_n)$converges to$s$, there exists$N$s.t.$\forall k\geq N$,$|s_n-s| \leq \frac{2(L^+-s)}{3}$, and thus$\forall k\geq N$,$s_k \leq s+\frac{2(L^+-s)}{3} <L^+ - \frac{L^+-s}{3}$, which means$s^+_N = \sup_{k\geq N} (s_k) \leq L^+- \frac{L^+-s}{3}$. But$L^+:= \inf_n s^+_n$, so$L^+ \leq s^+_N$, then$L^+ \leq L^+- \frac{L^+-s}{3}$, a contradiction. 1. If$\lim \sup_{n\to\infty} s_n=\lim \inf_{n\to\infty} s_n =s\in\mathbb{R}$, then the sequence$(s_n)$converges to$s$. Proof: Again we observe that we also have$L^+=\lim_{n\to\infty}(s^+_n)=s$and$L^-=\lim_{n\to\infty}(s^-_n)=s$. Let arbitrary$\epsilon > 0$. Then$\exists N_1 \in \mathbb{N}$, .s.t.$\forall k \geq N_1, s^+_k -s \leq \epsilon$; similarly$\exists N_2 \in \mathbb{N}$, .s.t.$\forall k \geq N_2, s-s^-_k \leq \epsilon$. Let$N=\max(N_1,N_2$). Then$\forall k \geq N$,$s-\epsilon \leq s^-_k \leq s_k \leq s^+_k \leq s+\epsilon$, hence$|s_k - s| \leq \epsilon\$.