Proof that $\lim \sup_n (a_n + b_n) \in \mathbb{R}$ when $\lim \sup_n a_n \in \mathbb{R}$, $\lim \sup_n b_n \in \mathbb{R}$

Question

I'm stuck in the middle of solving the following question.

Say $\lim \sup_n a_n = \alpha \in \mathbb{R}$, $\lim \sup_n b_n = \beta \in \mathbb{R}$. Then can we say the following?

$$ \lim \sup_n (a_n + b_n) \in \mathbb{R} $$

My attempt : define $x_n = \sup \{a_k : k \ge n \} $, $y_n = \sup \{b_k : k \ge n \} $. Then by the assumption, both $\{x_n : n \in \mathbb{N}\}$ and $\{y_n : n \in \mathbb{N}\}$ are bounded below.

Define $z_n = \sup \{a_k + b_k : k \ge n \}$. I think the fact I can use now is $z_n \le x_n + y_n$. But I find that this does not exclude the possibility of $\{z_n : n \in \mathbb{N}\}$ being NOT bounded below, i.e. $\lim_n z_n = -\infty$.

Is there anyone to help me proceed my proof? Or, can someone give a counter-example?

Answer 1

That's false.

Let $a_n = -n$ for even $n$, and $0$ otherwise.

Let $b_n = -n$ for odd $n$, and $0$ otherwise.

Then: $$\limsup_n a_n = 0,\,\limsup_n b_n = 0,\,\limsup_n (a_n+b_n) = -\infty$$