real analysis - Show that certain lim sup additivity implies convergence.

I'm trying to prove the following:

Let $\{a_n\}$ be a bounded sequence. If for every bounded sequence $\{b_n\}$ the following holds:

$$1) \limsup_{n \to \infty} (a_n + b_n) = \limsup_{n \to \infty} a_n + \limsup_{n \to \infty} b_n$$

Show that $\{a_n\}$ is convergent.

Outline of Proof:

Here's the outline of my proof. But there's a point I get stuck below. I will be using the following theorem:

2) Let $\{a_n\}$ be a bounded sequence, let $L = \limsup_{n \to \infty} a_n$. If $\epsilon > 0$ there exists infinitely many $n$ such that $a_n > L - \epsilon$, and there exists $N_1 \in \mathbb{N}$ such that if $n \ge N_1$ then $a_n < L + \epsilon$

Since condition $1)$ holds for every bounded sequence $\{b_n\}$ we can assume it holds for a $\{b_n\}$ that's convergent. That means $\forall \epsilon > 0$ $\exists N_2 \in \mathbb{N}$ $\ni$ if $n \ge N_2$ then $L_b - \epsilon < b_n < L_b + \epsilon$ where $L_b = \lim_{n \to \infty} b_n$.

Let $\epsilon > 0$, $L_a = \limsup_{n \to \infty} a_n$, $L_b = \limsup_{n \to \infty} b_n$. Let N = $\max\{N_1, N_2\}$.

If $n \ge N$ we can easily show $a_n < L_a + \epsilon$.

This is the point where I'm stuck. For infinitely many $n$:

$$a_n + b_n > L_a + L_b - \epsilon/2$$
$$a_n > L_a + L_b - b_n - \epsilon/2$$

$\forall n \ge N$: $L_b - b_n > -\epsilon/2$.

But the only conclusion I can draw from the above is that for infinitely many $n$ (and not $\forall n \ge N$):

$$a_n > L_a - \epsilon$$

Answer

Prove that
$$\limsup(-a_n) = -\liminf(a_n).$$
Then with setting $b_n := -a_n$ you are done.