# Prove the limit superior of a bounded sequence converges

Let $$(a_n)_{n=1}^\infty$$ be a bounded sequence and $$b_n = \sup\{a_k\ |\ k \geq n\}$$. Prove $$b_n$$ converges. This is the limit superior of $$(a_n) := \limsup\ a_n$$.

Wanted to see if my proof made sense.

• Since $$a_n$$ is bounded, we know $$\exists\ M \in \mathbb{R}$$ such that $$|a_n| \leq M\ \forall n \in \mathbb{N}$$.

• $$(b_n)$$ converges to $$x \in \mathbb{R}$$ if $$\forall \epsilon > 0, \exists\ N \in\ \mathbb{N}\ \forall n \in \mathbb{N}, n \geq N \implies |b_n-x| < \epsilon$$.

Since $$(a_n)$$ is bounded, by completeness of $$\mathbb{R}$$ and non empty by assumption, we know that since it has an upper bound, it has a least upperbound, that is the supremum.

So we know past some $$K$$, $$a_k$$ = $$S$$, where $$S$$ denotes the supremum of all elements in the sequence. So for sequence is eventually constant, that is constant for $$n \geq K$$. And so converges to $$S$$.

• Hint: the sequence $(b_n)_{n \in \mathbb{N}}$ is increasing and bounded. Your conclusion on the $a_k$ is false -- just take $a_{k} = 1/k$ for example. Oct 18, 2018 at 22:31
• Increasing and bounded means we can use the monotone convergence theorem. So follows from there I suppose.
– SS'
Oct 18, 2018 at 22:33
• @weirdo $b_n$ is decreasing of course.
– user
Oct 18, 2018 at 22:39
• ^ Was about to note. If we were referring to the limit inferior, then it'd be increasing as I understand. But decreasing for this case.
– SS'
Oct 18, 2018 at 22:40
• @SS' Yes for the liminf $I_n$ is increasing and the same argument applies.
– user
Oct 18, 2018 at 22:41

• $$b_n$$ is bounded, since $$a_n$$ is bounded
• $$b_n$$ is decreasing since by the definition $$b_{n+1}\le b_n$$