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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 $S$.

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    $\begingroup$ 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. $\endgroup$
    – weirdo
    Oct 18, 2018 at 22:31
  • $\begingroup$ Increasing and bounded means we can use the monotone convergence theorem. So follows from there I suppose. $\endgroup$
    – SS'
    Oct 18, 2018 at 22:33
  • $\begingroup$ @weirdo $b_n$ is decreasing of course. $\endgroup$
    – user
    Oct 18, 2018 at 22:39
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    $\begingroup$ ^ 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. $\endgroup$
    – SS'
    Oct 18, 2018 at 22:40
  • $\begingroup$ @SS' Yes for the liminf $I_n$ is increasing and the same argument applies. $\endgroup$
    – user
    Oct 18, 2018 at 22:41

1 Answer 1

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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.

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