# Why limit superior of bounded monotonic sequence exists and what it is?

I'm taking real analysis and I still find this concept hard to understand:

If S is any bounded nonempty set of real numbers, $$\sup(S)$$ and $$\inf(S)$$ exist.

However, consider a bounded decreasing sequence $$\{a_n\}$$. Take $$a_n= \frac 1 n$$). Consider $$\{b_n\}$$ given by $$b_n= \sup \{a_k: k \ge n \}$$. We could prove $$b_n$$ was monotonic and bounded, thus converges. We wrote $$\limsup a_n =\lim b_n$$. In our case, $$\lim b_n = \lim \frac 1 n = 0$$.

However, $$b_n$$ is the superior of $$\{a_n\}$$, as the least upper bound of $$a_n$$, thus $$b_n$$ must be bigger or equal to $$a_n$$, which contradicts the result.

I searched online for limit superior but couldn't understand the notation well.

Could you help me to explain:

1. What is the real definition of limit superior in $$\mathbb{R}$$?
2. What went wrong, and the correct answer to the question?
• $\lim\!\sup$ is not the same as $\sup$. Feb 15, 2017 at 1:27
• In your Definition, you have $a_n = b_n$ as $a_k$ is decreasing. Feb 15, 2017 at 1:29

The real definition of $\limsup$ is the one you wrote down: $$\limsup_{n\to\infty} a_n = \lim_{n\to\infty} b_n$$ where $b_n=\sup(\{a_k|k\ge n\})$
Why does $b_n$ being the least upper bound of the tail cause a contradiction? It's certainly true that $b_n\ge a_n$ but I don't see how that's a problem.
It's totally possible and commonplace for the limit (and limsup) of a sequence to be greater than any element of the sequence. For instance $\lim_n (1-1/n)= \limsup_n(1-1/n)=1,$ and we have $1>a_n$ for all $n$.
Also, addressing the title of your question, I think you may be confused. The limsup of a sequence always exists (in $[-\infty,\infty]$). No need for boundedness or monotonicity. Where monotonicity comes in is in the proof of this. $b_n,$ as defined above, is a monotonically decreasing sequence in $[-\infty,\infty]$, and thus, by a theorem, always has a limit in $[-\infty,\infty].$ If the sequence $a_n$ is bounded (i.e. there is a $B\in \mathbb R$ such that $|a_n|<B$ for all $n$), then $b_n$ is a monotonically decreasing bounded sequence in $\mathbb R$ and thus by another version of the same theorem, the limit exists in $\mathbb R.$
So the only way your title makes sense to me is if you're trying to prove that the limsup exists in $\mathbb R$ (as opposed to $[-\infty,\infty]$) and in that case you only need boundedness of $a_n,$ not monotonicity.