TITLE: Prove $\limsup\limits_{n \to \infty} (a_n+b_n) \le \limsup\limits_{n \to \infty} a_n + \limsup\limits_{n \to \infty} b_n$
QUESTION [47 upvotes]: I am stuck with the following problem.

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

I was thinking of using the triangle inequality saying $$|a_n + b_n| \le |a_n| + |b_n|$$ but the problem is not about absolute values of the sequence.
Intuitively it's clear that this is true because $a_n$ and $b_n$ can "reduce each others magnitude" if they have opposite signs, but I cannot express that algebraically...
Can someone help me out ?

REPLY [2 votes]: Because the sup of a set is no less than the sup of a subset 
$$
\begin{align}
\sup_{k\ge n}a_k+\sup_{k\ge n}b_k
&=\sup_{\substack{j\ge n\\k\ge n}}(a_j+b_k)\\
&\ge\sup_{\substack{j\ge n\\k\ge n\\j=k}}(a_j+b_k)\\
&=\sup_{k\ge n}(a_k+b_k)\tag1
\end{align}
$$
Since $\sup\limits_{k\ge n}a_k$ is non-increasing, $\lim\limits_{n\to\infty}\sup\limits_{k\ge n}a_k$ exists. Taking $\lim\limits_{n\to\infty}$ of $(1)$ gives
$$
\limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n\ge\limsup_{n\to\infty}(a_n+b_n)\tag2
$$