Proving $\lim \sup (c a_n) = c \cdot \lim \sup a_n$

Question

Let $(a_n)$ be a bounded sequence. Let $c \in \mathbb{R}$ and suppose that $c \geq 0$. Prove then that $$ \lim_{n \to \infty} \sup (ca_n) = c \cdot \lim_{n \to \infty} \sup a_n. $$

Attempt at proof: I managed to prove one inequality. Here is my reasoning:

$(a_n)$ is a bounded sequence. For every $n \in \mathbb{N}$ the subset $A_n = \left\{ a_m \mid m \geq n \right\}$ is then a bounded subset of $\mathbb{R}$. Hence it has a supremum. Let $y_n = \sup A_n$. Now consider the set $$B_n = \left\{ c a_m \mid m \geq n \right\}. $$ Let $ m \geq n$ be arbitrary. Because $a_m \in A_n$, we have that $a_m \leq y_n$. Because $c \geq 0$, it also holds that $ca_m \leq c y_n$. But $ca_m \in B_n$. Since $m$ was chosen arbitrarily, this proves that $c y_n$ is an upper bound for $B_n$. But $\sup(B_n)$ is the least upper bound. Hence it follows that $\sup(B_n) \leq c y_n$. By taking the limit it now follows that $$ \lim_{n \to \infty} \sup (ca_n) \leq c \cdot \lim_{n \to \infty} \sup a_n.$$ But I don't know how to prove the other inequality. I tried to assume that $$ \lim_{n \to \infty} \sup (ca_n) < c \cdot \lim_{n \to \infty} \sup a_n $$ and then to derive a contradiction, but I was not able to.

Any help please to proof the other inequality?

Answer 1

If $c=0$ the conclusion is quite obvious. We need only consider $c > 0$.

When proving things like this, you often prove one inequality first and then simply use that inequality to prove the other. In your case, you've already proven \begin{equation} \tag{1} \limsup_{n\to \infty} (ca_n) \le c\limsup(a_n). \end{equation} Now consider $$\limsup_{n\to \infty} a_n = \limsup_{n\to \infty} \left(\frac 1 c c a_n \right) \le \frac 1 c \limsup_{n\to \infty} (ca_n) \,\,\,\,\,\,\,\,\,\, \text{ by } (1).$$ Multiplying both sides by $c$ gives $$c \limsup_{n\to \infty} a_n \le \limsup_{n\to \infty} (ca_n)$$ which is the desired conclusion.