# If $\lim\sup a_n = -\infty$, then $a_n\to -\infty$, and if $\lim\sup a_n = \infty$ there exists a subsequence of $a_n$ that $\to\infty$

Show that if $$\lim\sup a_n = -\infty$$, then $$a_n\to -\infty$$, and if $$\lim\sup a_n = \infty$$ there exists a subsequence of $$a_n$$ that $$\to\infty$$. What if $$\lim\inf a_n = \pm\infty$$? We're working with the extended real number system, $$\bar{\mathbb{R}} = \mathbb{R} \cup\{\infty,-\infty\}$$.

My work: $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m = -\infty$$ is known. This means that there exists some $$n'\in\mathbb{N}$$ such that $$\sup_{m\ge n'}a_m = -\infty$$. So, for all $$m \ge n'$$, we have $$a_m \le -\infty$$. This can happen iff $$a_m = -\infty$$ for all $$m\ge n'$$. Hence, $$a_n\to -\infty$$.

Is this proof alright? I'm concerned about steps where I've written $$a_m=-\infty$$ and such, even though we know that we are working with the extended real number system. Is this okay to do? Just some uneasiness since I haven't used $$=$$ with infinities before.

Now assume that $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}$$ I can intuitively see that the least upper bound of all tail sequences of $$a_n$$ is $$\infty$$, so maybe $$a_n\to\infty$$ itself? I'm not sure because the problem explicitly asks to produce a subsequence of $$a_n$$ (or show that one exists) that goes to $$-\infty$$. How do we do this?

Similarly for the case when $$\lim\inf a_n = \pm \infty$$, what can we say? I'd guess that $$\lim\inf a_n = \infty$$, then $$a_n\to\infty$$ and if $$\lim\inf a_n = -\infty$$ then there is a subsequence of $$a_n$$ that goes to $$\infty$$.

Here's my definition for diverging to $$+\infty$$: $$x_n\to +\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n > c)$$ $$x_n\to -\infty:= \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n > m\ (x_n < c)$$

## Second Attempt:

$$\lim\sup_{n\to\infty} a_n = -\infty \implies \lim_{n\to\infty} \sup\{a_k:k\ge n\} = -\infty$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (\sup\{a_k:k\ge n\} < c)$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_k < c, k\ge n)$$ $$\implies \forall c\in\mathbb{R}\ \exists m\in\mathbb{N}\ \forall n>m\ (a_n < c) \implies \lim_{n\to\infty} a_n = -\infty$$ Similarly, $$\liminf_{n\to\infty} a_n = \infty \implies a_n\to\infty$$

• I think the definition for diverging to $+\infty$ is that for any $M\in\mathbb R$, there exists an $N\in\mathbb N$ so that $n>N$ implies that $a_n>M$. (Similar for diverging to $-\infty$.) Commented Jan 19, 2021 at 6:54
• What does $a_m\leq-\infty$ mean for a real number? Commented Jan 19, 2021 at 9:07
• @MichaelBurr $a_m = -\infty$ perhaps? There is nothing smaller than $-\infty$ so Commented Jan 19, 2021 at 9:11
• Commented Jan 19, 2021 at 14:10

No, your proof is not correct. There exists in general no $$n'\in\mathbb{N}$$ such that $$\sup_{m\ge n'}a_m = -\infty$$, but there exists a subsequence $$(n_k)$$ so that $$a_{n_k}\to-\infty(k\to\infty)$$ and $$\sup_{m\ge n}a_m \to -\infty (n\to\infty)$$. The first comes from the fact that $$\limsup a_n=c$$ means your biggest cluster point is $$c$$ and cluster points are exactly the limits of converging subsequences. So $$\limsup$$ is the biggest extended real number a subsequence converges. The second can either be implied by the first or you can use $$\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{m\geq n} a_m$$, which comes from the fact that $$(\sup_{m\geq n} a_m)_{n\in\mathbb N}$$ is a decreasing sequence so you can exchange $$\inf$$ and $$\lim$$ in $$\limsup_{n\to\infty} a_n=\inf_{n\in\mathbb N}\sup_{m\geq n} a_m=\lim_{n\to\infty}\sup_{m\geq n}a_m.$$

What you have to show (as boink already mentioned in the comments) is that for every $$M\in\mathbb R$$ (especially for big negative $$M$$) exists a $$N\in\mathbb N$$ so that for all $$n\geq N$$ holds $$a_n. So choose any $$M$$ and now let's find such a $$N$$. Your first step to consider the sequence $$(\sup_{m\ge n}a_m)_{n\in\mathbb N}$$ was in the right direction. As already mentioned above we have $$\sup_{m\ge n}a_m \to -\infty (n\to\infty)$$. This means we find a $$N'$$ so that for every $$n\geq N'$$ holds $$\sup_{m\ge n}a_m . This implies that for every $$n\geq N'$$ holds $$a_n. This number is the $$N=N'$$ we were looking for and we have finished the proof.

Now consider $$\lim\sup a_n = \inf_{n\ge1}\sup_{m\ge n} a_m = \infty \implies \sup_{m\ge n} a_m = \infty \text{ for all }n\in\mathbb{N}.$$ The $$\lim\sup$$ means the bigest cluster point of your sequence is $$\infty$$, which directly implies the statement, since you find for an arbitrary big $$n$$ a $$m>n$$ so that $$a_m$$ is arbitrary big. If you don't wanna handle with cluster points here, you can use the fact from above $$\infty=\limsup_{n\to\infty} a_n=\lim_{n\to\infty}\sup_{m\geq n}a_m$$ and then apply that the decreasing sequence $$(\sup_{m\ge n}a_m)_{n\in\mathbb N}$$ converges to $$\infty$$ which means it already has to be $$\infty$$.

Your last two guess regarding $$\lim\inf a_n = \pm \infty$$ are correct.

What you wrote under "Second Attempt" is comletely correct.

To give you an alternative proof to the first part, you could also use the property $$\lim\inf a_n\leq \lim\sup a_n,\tag{1}\label{tag}$$ where equality holds if and only if the limit $$\lim a_n$$ exists and $$\lim\inf a_n= \lim\sup a_n= \lim a_n.$$ So if $$\lim\sup a_n=-\infty$$ already holds, it follows form \eqref{tag} that $$\lim\inf a_n=-\infty$$ and the statement follows from the 'iff'-part.

• Thanks for your answer. Could you first clarify if $\lim_{n\to\infty} \sup_{m\ge n} a_m = \inf_{n\ge1}\sup_{m\ge n} a_m$? Are they the same thing? Commented Jan 19, 2021 at 9:17
• Yes, your first comment is correct.
– mag
Commented Jan 19, 2021 at 9:20
• $\lim\sup a_n=c$ means your biggest cluster point is $c$. But cluster points are exactly the limits of converging subsequences. So $\lim\sup$ is the biggest extended real number a subsequence converges.
– mag
Commented Jan 19, 2021 at 9:24
• In that case, I'm not able to see where your first statement is coming from. Is there a proof for the fact that $\lim\sup a_n = \pm\infty$ implies that there exists a subsequence $a_{n_k}$ such that $\sup_{m\ge n_k} \to \pm\infty$? Commented Jan 19, 2021 at 10:12
• Sry, I made something a bit confusing. It is not only a subsequence $(n_k)$ sucht that $\sup_{m\ge n_k}a_m\to-\infty$, but it holds for the complete sequence $\sup_{m\ge n}a_m\to-\infty$. The 'subsequence'-part belongs to the actual sequence $a_{n_k}$. I corrected it in the answer.
– mag
Commented Jan 19, 2021 at 10:40

The statement $$\sup_{m\geq n'}\alpha_m =-\infty,$$ for some $$n'$$ is false in general. For example if $$\alpha_m = -m$$ then for every $$n'$$ we have $$\sup_{m\geq n'}=\alpha_{n'}=-n'>-\infty$$ and $$\alpha_n\to -\infty$$ as $$n\to \infty$$. To prove that $$\alpha_n\to -\infty$$ when $$\limsup_{n}\alpha_n = -\infty$$ fix some $$c\in\mathbb{R}$$. Since $$\limsup_n \alpha_n = - \infty$$ there is $$n'$$ such that $$\sup_{m\geq n'}\alpha_m < c.$$ This means that for every $$m\geq n'$$ we have $$\alpha_m which is equivalent to $$\alpha_n\to -\infty$$. Now, if $$\limsup_{n}\alpha_n =\infty$$ it does not necessarily implies that $$\alpha_n \to \infty$$. For example consider the sequence $$\alpha_n=\begin{cases} n,\ n \text{ is even }\\ -1,\ n \text{ is odd} \end{cases}$$ Then, the subsequence $$\alpha_{2n-1}$$ is constant and equal to $$-1$$, hence $$\alpha_n$$ cannot converge to $$\infty$$. Although, it is true that there exists a subsequence $$\alpha_{k_n}$$ such that $$\alpha_{k_n}\to \infty$$. Since $$\limsup_n \alpha_n =\infty$$ for every $$n$$ you have that $$\sup_{m\geq n}\alpha_m =\infty$$. In particular, $$\sup_{m\geq 1 }\alpha_m >1.$$ This means, that there exists some $$k_1\geq 1$$ such that $$\alpha_{k_1}>1$$. Now, using again the fact that $$\sup_{m\geq k_1 +1}\alpha_m>2$$ we may find $$k_2 >k_1$$ such that $$\alpha_{k_2}>2$$. Continuing this way, recursively, we may find indices $$k_1 in ascending order such that $$\alpha_{k_n}>n$$ for every $$n$$. This means that the subsequence $$\alpha_{k_n}$$ converges to $$\infty$$. Similarly, if $$\liminf_n \alpha_n=\infty$$ then $$\alpha_n\to \infty$$ and if $$\liminf_n \alpha_n =-\infty$$ then there exists a subsequence $$\alpha_{k_n}\to -\infty$$.