Prove that $\limsup_{n \to \infty} (a_n+b_n)= a + \limsup_{n \to \infty} b_n$

Question

I have to prove the following:

Given that lim $a_n$ exists and that lim $a_n$ = $a\in \mathbb{R}$ prove that: $$\limsup_{n \to \infty} (a_n+b_n)= a + \limsup_{n \to \infty} b_n$$

I proved one side of the inequality that is, I showed that $$\limsup_{n \to \infty} (a_n+b_n) \le a + \limsup_{n \to \infty} b_n$$ by using supremums of $(a_n)$, $(b_n)$ and $(c_n)$ where $(c_n)$={$c_k$|$k$$\ge$$n$} and $c_k=a_k+b_k$. But I can't figure out how to prove the converse. Can anyone explain how to do that without using $\epsilon$ or subsequences ? Any help would be appreciated thanks.

Answer 1

Since

\begin{align}\limsup_{n\to \infty} (a_n + b_n) - \liminf_{n\to \infty} a_n &= \limsup_{n\to \infty} (a_n + b_n) + \limsup_{n\to \infty} (-a_n)\\ &= \lim_{n\to \infty} [\sup_{k\ge n} (a_k + b_k) + \sup_{k\ge n} (-a_k)]\\ &\ge \lim_{n\to \infty} \sup_{k\ge n}[(a_k + b_k) + (-a_k)] \\ &= \lim_{n\to \infty} \sup_{k\ge n} b_k\\ &= \limsup_{n\to \infty} b_n, \end{align}

we have

$$\limsup_{n\to \infty} (a_n + b_n) \ge \liminf_{n\to \infty} a_n + \limsup_{n\to \infty} b_n = a + \limsup_{n\to \infty} b_n.$$