Suppose {$a_n$} and {$b_n$} are bounded sequences and that lim $b_n =b$. Prove that $$\lim \sup (a_n + b_n) = \lim \sup a_n + b. $$
Here is what I tried:
Consider $$\sup (a_k) + b$$ Since $b$ is a fixed number, and $\sup (a_k) = \sup \{ a_n : n\geq k \} $, then it follows that $$\sup (a_k) + b = \sup (a_k + b)$$ (is this correct?)
Then since $b_k \rightarrow b $, for $k \geq N$ for some $N \in \mathbb{N}$ large enough, $ b_k - \epsilon \leq b \leq b_k + \epsilon$. So $$ \sup (a_k + b_k - \epsilon) \leq \sup (a_k + b) \leq \sup (a_k + b_k + \epsilon)$$
Since this holds for all $k \geq N$, we can take the limit as $k$ approaches $\infty$, $$\lim \sup (a_k + b_k - \epsilon) \leq \lim \sup (a_k + b) \leq \lim \sup (a_k + b_k + \epsilon)$$
Since $\epsilon$ was arbitrary, we can conclude that $$\lim \sup (a_n + b_n) = \lim \sup a_n + b. $$
To do it from scratch, go back to the definition of $\limsup$. Set $g_k=\sup_{n\ge k}\{a_n\}$ and $h_k=\sup_{n\ge k}\{b_n\}.$ Then, for all $n\ge k,$
$a_n+b_n\le g_k+h_k$ so $\sup_{n\ge k}(a_n+b_n)\le g_k+h_k=\sup_{n\ge k}\{a_n\}+\sup_{n\ge k}\{b_n\}.$ That is,
$\tag1 \sup_{n\ge k}(a_n+b_n)\le \sup_{n\ge k}\{a_n\}+\sup_{n\ge k}\{b_n\}.$
So, in our case
$\tag2 \underset{k\to \infty}\lim\sup_{n\ge k}(a_n+b_n)\le \underset{k\to \infty}\lim\sup_{n\ge k}\{a_n\}+\underset{k\to \infty}\lim\sup_{n\ge k}\{b_n\}=\underset{k\to \infty}\lim\sup_{n\ge k}\{a_n\}+b,$
which is of course
$\tag3 \limsup (a_n + b_n)\le \limsup a_n+b$
For the reverse inequality, we use a trick: $a_n = (a_n + b_n) + (-b_n)$ and now $(3)$ gives
$\tag4 \limsup a_n\le \limsup (a_n + b_n)-b\Rightarrow \limsup a_n+b\le \limsup (a_n + b_n).$