Exercise 2.4.7

Question

[Limit Superior]
  \label{ex:lim_sup}
  Let $\left(a_{n}\right)$ be a bounded sequence.

\begin{enumerate}[label=(\alph*)] 
  \item Prove that the sequence defined by $y_{n}=\sup \left\{a_{k}: k \geq n\right\}$ converges.
  \item The limit superior of $\left(a_{n}\right)$, or $\lim \sup a_{n}$, is defined by
    $$
    \limsup a_{n}=\lim y_{n}
    $$
    where $y_{n}$ is the sequence from part (a) of this exercise. Provide a reasonable definition for $\lim \inf a_{n}$ and briefly explain why it always exists for any bounded sequence.
  \item Prove that $\lim \inf a_{n} \leq \lim \sup a_{n}$ for every bounded sequence, and give an example of a sequence for which the inequality is strict.
  \item Show that $\lim \inf a_{n}=\lim \sup a_{n}$ if and only if $\lim a_{n}$ exists. In this case, all three share the same value.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item $(y_n)$ is decreasing and converges by the monotone convergence theorem.
  \item Define $\lim \inf a_n = \lim z_n$ for $z_n = \inf\{a_n : k \ge n\}$. $z_n$ converges since it is increasing and bounded.
  \item Obviously $\inf\{a_k : k \ge n\} \le \sup\{a_n : k \ge n\}$ so by the Order Limit Theorem $\lim \inf a_n \le \lim \sup a_n$.
  \item If $\lim \inf a_n = \lim \sup a_n$ then the squeeze theorem (Exercise 2.3.3) implies $a_n$ converges to the same value, since $\inf\{a_{k\ge n}\} \le a_n \le \sup\{a_{k\ge n}\}$.
   \end{enumerate}

Exercise 2.4.7

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Exercise 2.4.7

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