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

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?

## 1 Answer

That's false.

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

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

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