Equality for limit superior: $\limsup_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$

Question

I'm trying to prove the following equality regarding the limit superior:

Let $(a_n)_{n\in\mathbb{N}}$ be a convergent sequence and let $(b_n)_{n\in\mathbb{N}}$ be bounded. Then

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

I worked out some things but I'm unsure whether they are correct:

I've already prove a theorem stating that $\limsup_{n\to\infty}(a_n+b_n)\leq\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$ for general $(a_n),(b_n)$. As $(a_n)$ is convergent, the $\limsup_{n\to\infty}a_n$ replaces with a $\lim_{n\to\infty}a_n$ and what remains to show is that $\limsup_{n\to\infty}(a_n+b_n)\geq\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n$ for a convergent $(a_n)$.

Writing the definition of the limit superior, we have that $\limsup_{n\to\infty}(a_n+b_n)=\lim_{n\to\infty}\sup\{a_n+b_n,\dots\}$. Now for this supremum expression, we have that $\sup\{a_n+b_n,\dots\}\geq\sup\{a+b_n,\dots\}$ as $(a_n)$ converges to $a$ and the sequence of suprema should be decreasing.

By the laws of the supremum, we then have $\sup\{a+b_n,\dots\}=a+\sup\{b_n,\dots\}$. Running n to infinity should yield $\lim_{n\to\infty}a+\sup\{b_n,\dots\}=a+\limsup_{n\to\infty}b_n$ and thus $\limsup_{n\to\infty}(a_n+b_n)\geq a+\limsup_{n\to\infty}b_n$.

Answer 1

HINT: Note that $$\limsup b_n=\limsup (b_n+a_n+(-a_n))\leq\limsup (b_n+a_n)+\limsup (-a_n).$$ Now, use that $$\limsup (-a_n)=-\liminf a_n=-\lim a_n.$$

Answer 2

$\sup\{a_{n}+b_{n},...\}\geq\sup\{a+b_{n},...\}$ seems not correct for general, but you are close to the goal. My way: Let $\epsilon>0$, there exists some $N$, for $n\geq N$, $a-\epsilon<a_{n}$, so $a-\epsilon+b_{n}<\sup\{a_{n}+b_{n},...\}$, take $\sup$ to $n\geq N$, then $a-\epsilon+\sup_{n\geq N}\{b_{n}\}\leq\sup\{a_{n}+b_{n},...\}$, but $\limsup_{n} b_{n}\leq\sup_{n\geq N}\{b_{n}\}$, so $a-\epsilon+\limsup_{n}b_{n}\leq\sup\{a_{n}+b_{n},...\}$, $\epsilon>0$ is arbitrarily, the result follows.