If $\limsup\limits_{n\rightarrow \infty} a_n=a< \infty$, then there exists for all $\varepsilon >0$ a $N\in \mathbb{N}$ such that $a_n\leq a+\varepsilon$ for all $n\in \mathbb{N}$, $n\geq N$

My attempt:

Suppose there are infinite elements $a_{n_1},a_{n_2}, a_{n_3},...$ with $a_{n_k}\geq a$. The sequence $(a_{n_k})_k$ is bounded above, otherwise $(a_n)_n$ wouldn't be bounded and $\limsup\limits_{n\rightarrow \infty} a_n=+\infty$. Hence $(a_{n_k})_k$ has - according to Weierstraß and Bolzano - a convergent subsequence $(a_{n_{k_j}})$ with a limit $\geq a$} since $(a_{n_{k_j}})\geq a$ for every $j\in \mathbb{N}$.

  • $\begingroup$ How do you define $\limsup_na_n$? $\endgroup$ – José Carlos Santos Jun 2 at 12:03
  • $\begingroup$ Like that: $\limsup\limits_{n\rightarrow \infty}a_n:=\begin{cases}\sup H, \text{ if } (a_n)_n \text{is bounded above} \\ \infty, \text{ else}\end{cases}$. $H$ is the set of limit points. $\endgroup$ – ParabolicAlcoholic Jun 2 at 12:06

You don't need to go as far as using Bolzano-Weierstraß theorem. We can prove your statement directly.

Recall that $\limsup\limits_{n\rightarrow \infty} a_n = \lim\limits_{n\to\infty} \sup\limits_{k\geq n} a_k$. Let us fix $\epsilon > 0$. By definition of the limit, we know that there is some $N\in \mathbb N$ such that $\sup\limits_{k\geq N} a_k < a+\epsilon$. Thus, by definition of the supremum, we conclude that $a_k < a + \epsilon$ for every $k\geq N$, which is the conclusion you wanted.


Suppose otherwise. That is, suppose that, for each $N\in\mathbb N$, there is a $n\in\mathbb N$ such that $n\geqslant N$ and that $a_n\geqslant a+\varepsilon$. So, there is a sequence $(a_{n_k})_{k\in\mathbb N}$ such that $(\forall k\in\mathbb N):a_{n_k}\geqslant a+\varepsilon$. And, since $(a_n)_{n\in\mathbb N}$ is bounded, $(a_{n_k})_{k\in\mathbb N}$ is bounded too. So, it has a convergent subsequence, by the Bolzano-Weierstrass theorem. The limit of this subsequence must be greater than or equal to $a+\varepsilon$, but is is impossible, since, by definition, $a$ is the supremum of the set of limit points.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.