- #1

Bashyboy

- 1,421

- 5

## Homework Statement

The limit superior of a sequence ##(a_n)## is defined as ##\overline{\lim}_{n \to \infty} a_n = \lim_{n \to \infty} \sup \{a_k ~|~ k \ge n \}##. Letting ##L = \overline{\lim}_{n \to \infty} a_n##, I am asked to prove the following:

(i) For each ##N## and for each ##\epsilon > 0##, there exists some ##k > N## such that ##a_k \ge L - \epsilon##.

(ii) For each ##\epsilon > 0##, there is some ##N## such that ##a_k \le L + \epsilon## for all ##k > N##.

(iii) ##\overline{\lim}_{n \to \infty} ca_n = c \overline{\lim}_{n \to \infty} a_n##

## Homework Equations

## The Attempt at a Solution

(iii) is rather easy to prove: it just follows from the fact that ##\sup(cA) = c \sup (A)## and ##\lim_{n \to \infty} c x_n = c \lim_{n \to \infty} x_n## for ##c \ge 0##, so I will move on to the first and second part

Let ##\epsilon > 0## and ##N \in \Bbb{N}##. Since ##L = \overline{\lim}_{n \to \infty} a_n##, then there exists a ##K \in \Bbb{N}## such that ##| \sup \{a_k ~|~ k \ge n \} - L | < \epsilon## for every ##n \ge K## or ##L - \epsilon < \sup \{a_k ~|~ k \ge n \} < \epsilon + L## for every ##n \ge K##.. On the one hand, we get ##a_k \le \sup \{a_k ~|~ k \ge n \} < \epsilon + L## for every ##k \ge K##, which proves part (ii). However, I don't see how to prove part (i). Also, as you may have noticed, I can only get strict inequality in part (ii).