Prove the limit superior of a bounded sequence converges

Question

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 $S$.

Answer 1

We have that

• $b_n$ is bounded, since $a_n$ is bounded

and

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

then refer to Monotone convergence theorem.