# ($s_n$) bounded above implies lim sup ($s_n$) is finite?

I'm reading Elementary Analysis by Ross and working through the proof of Theorem 11.7 (see below). I don't understand the claim in the proof that "... ($$s_n$$) is bounded above, so that lim sup $$s_n$$ is finite." For example, consider the sequence {$$-n^2$$} for all $$n \in \mathbb{N}$$. This is bounded above by $$0$$, but isn't lim sup $$s_n$$ = $$-\infty$$ in this case, and therefore not finite? What am I missing?

See various theorem references and proof below from the text:

Theorem 11.7:

Let ($$s_n$$) be any sequence. There exists a monotonic subsequence whose limit is $$\limsup s_n$$ and there exists a monotonic subsequence whose limit is $$\lim inf s_n$$.

Proof:

If ($$s_n$$) is not bounded above, then by Theorem 11.2(ii), a monotonic subsequence of ($$s_n$$) has limit $$+\infty = \limsup s_n$$. Similarly, if ($$s_n$$) is not bounded below, a monotonic subsequence has limit $$-\infty = \liminf s_n$$.

The remaining cases are that ($$s_n$$) is bounded above or is bounded below. These cases are similar, so we only consider the case that ($$s_n$$) is bounded above, so that lim sup $$s_n$$ is finite. Let $$t = \limsup s_n$$, and consider $$\epsilon > 0$$. There exists $$N_0$$ so that

$$\sup\{s_n : n > N\} < t + \epsilon$$ for $$N \geq N_0$$.

In particular, $$s_n$$ $$< t + \epsilon$$ for all $$n > N_0$$. We now claim

$$\{n \in \mathbb{N} : t - \epsilon$$ < $$s_n$$ < $$t + \epsilon\}$$ is infinite.

Otherwise, there exists $$N_1 > N_0$$ so that $$s_n$$ $$\leq t - \epsilon$$ for $$n > N_1$$. Then $$\sup \{s_n : n > N\} \leq t - \epsilon$$ for $$N \geq N_1$$, so that $$\limsup s_n < t$$, a contradiction. Since (1) holds for each $$\epsilon > 0$$, Theorem 11.2(i) shows that a monotonic subsequence of $$(s_n)$$ converges to $$t = \limsup s_n$$.

Theorem 11.2(i):

If $$t$$ is in $$\mathbb{R}$$, then there is a subsequence of $$(s_n)$$ converging to $$t$$ if and only if the set $$\{n \in \mathbb{N} : |s_n - t| < \epsilon\}$$ is infinite for all $$\epsilon > 0$$.

Theorem 11.2 (ii):

If the sequence $$(s_n)$$ is unbounded above, it has a subsequence with limit $$+\infty$$.

• Alternatively, and this might be obvious, but I thought it was worth mentioning, if $(s_n)_{n\in\mathbb{N}}$ is a sequence, then $\limsup(s_n)$ and $\liminf(s_n)$ are finite if and only if $(s_n)_{n\in\mathbb{N}}$ is bounded above and below. Jun 4 at 9:36

You are correct that there is a slight error here. The proof should probably say "so that $$\limsup s_n<\infty$$; if $$\limsup s_n=-\infty$$ then also $$\liminf s_n=-\infty$$ and we already have a suitable sequence, so we may assume $$\limsup s_n$$ is finite."

An alternative would be to start with the observation that if $$\limsup s_n=\liminf s_n$$ then any monotonic subsequence works for both, so we may assume they are different.

Let me first state a lemma:

Lemma: If $$\limsup (x_n)$$ is finite then $$(x_n)$$ is bounded above.

Proof: Let $$s=\limsup (x_n)$$ and so there exists $$N\in \mathbb N$$ such that we have $$x_n\lt s+1$$ for all $$n\gt N$$. It follows that $$x_n\lt\max\{x_1,x_2,\cdots,x_N,s+1\}$$. This proves our lemma.

Now, coming back to your question. You are right about observing that $$\limsup (-n^2)=-\infty$$ and this shows that converse to the lemma stated above is not true, which you probably thought should be true.

• Could you please explain why $\limsup(-n^2) = -\infty$?
– fwd
Jun 4 at 11:25
• @fwd: Use definition of limsup. Every real no. bounds the sequence $(-n^2)$ from above when n is large enough so what's the lowest upper bound?
– Koro
Jun 4 at 12:09
• You're right, thanks. For some reason I was stupid and kept assuming only non-negative reals bound the sequence above.
– fwd
Jun 4 at 12:15
• @fwd: It's absolutely no problem. It's part of learning. Glad that you understand it now :)
– Koro
Jun 4 at 12:17