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 ?
Define for all natural numbers $k$: $A_k = \sup\{ a_n: n \ge k \}$, $B_k = \sup\{ b_n: n \ge k \}$ (where $A_k, B_k \in \mathbb{R} \cup \{+\infty\}$, (They are decreasing because for larger $k$ we take the $\sup$ of fewer terms), so that by definition $\limsup_{n \to \infty} a_n = \lim_{k \to \infty} A_k$ and similarly for $B_k$ and $\limsup_{n \to \infty} b_n$. Also we consider the $C_k = \sup \{ (a_n + b_n) : n \ge k \}$, so that $\lim_{k \to \infty} C_k = \limsup_{n \to \infty} (a_n+b_n)$.
Now, fix an index $k$, then for all $n \ge k$ we have $a_n + b_n \le A_k + B_k$, because we estimate $a_n$ by the supremum of all terms of $(a_n)$ with $n \ge k$ and likewise for the $b_n$. As (for fixed $k$) the right hand side is fixed:
$$C_k = \sup \{ (a_n + b_n : n \ge k \} \le A_k + B_k\mbox{.}$$
This holds for all $k$, so we take the $\inf$ or $\lim$ on both sides as $k$ tends to infinity, and this preserves the inequality and we are done.
I happened to making solutions for PMA, so I would like to share my solution here:
If either $\limsup_{n \to \infty} a_n = +\infty$ or $\limsup_{n \to \infty} b_n = +\infty$, there is nothing to prove. So we may assume $\limsup_{n \to \infty} a_n = A, \limsup_{n \to \infty} b_n = B$, where $A < +\infty, B < +\infty$ (but each of them can possibly take $-\infty$).
Given $\varepsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$, such that $a_n < A + \varepsilon/2$ for all $n \geq N_1$ and $b_n < B + \varepsilon/2$ for all $n \geq N_2$. Take $N = \max(N_1, N_2)$, it follows that for all $n \geq N$, $$a_n + b_n < A + \varepsilon/2 + B + \varepsilon/ 2 = A + B + \varepsilon.$$ Let $n \to \infty$ in the above equation, we conclude that $\limsup_{n \to \infty} (a_n + b_n) \leq A + B + \varepsilon$. Since $\varepsilon$ is arbitrary, it follows that $\limsup_{n \to \infty}(a_n + b_n) \leq A + B$, proving the result.
Hint:
Given two sequences, $\;\displaystyle \{a_n\}_{n \in \Bbb N},\;$ $\,\{b_n\}_{n \in \Bbb N},\;$
and given the definition of the supremum of a sequence, we can see that for every $k\geq n$, $$(a_k + b_k) \;\; \leq \;\;\sup_{k\geq n} a_k + \sup_{k\geq n} b_k\,.$$
Now how does this imply that $$\lim_{n\to \infty} \sup(a_n + b_n) \;\; \leq \;\; \lim _{n \to \infty} \sup a_n + \lim_{n\to \infty} \sup b_n\quad ?$$
Added: see Definition 3.16, Theorem 3.17, 3.19: perhaps you'd prefer to use the notation used there.
This question occurs in Rudin's PMA, page 78, Exercise 5. Below is my trial of proof. Is it right? Any comments are welcome!
5. $\quad $For any two real sequences $\{a_n\}, \{b_n\},$ prove that \begin{gather*} \limsup_{n\to\infty}(a_n+b_n)\leq \limsup_{n\to\infty} a_n+\limsup_{n\to\infty} b_n, \end{gather*} provided the sum on the right is not of the form $\infty-\infty.$
proof: $\quad$ Put $a^*=\limsup\limits_{n\to\infty} a_n, b^*=\limsup\limits_{n\to\infty} b_n$ and $c^*=\limsup\limits_{n\to\infty} (a_n+b_n).$ We shall show $c^*\leq a^*+b^*.$
Since it is excluded that the right is of the form $\infty-\infty,$ if one of $a^*$ and $ b^*$ is $-\infty,$ then the assertion holds. So we assume that $a^*, b^*>-\infty.$ And if one of $a^*$ or $b^*$ is $+\infty,$ then $c^*=a^*+b^*,$ so we we need to consider the case $a^*, b^*\in\mathbb{R}.$
By Theorem 3.17 (see Page 56 of Rudin's PMA), for every $\epsilon>0$ there exists $N_1, N_2$ such that \begin{gather*} a_n<a^*+\frac{\epsilon}{2},\qquad \forall n>N_1,\\ b_n<b^*+\frac{\epsilon}{2},\qquad \forall n>N_2. \end{gather*} Hence we have \begin{gather*} a_n+b_n<a^*+b^*+\epsilon,\qquad \forall n>\max\{N_1, N_2\}. \end{gather*} Then, for every convergent subsequence $a_{\sigma(n)}+b_{\sigma(n)}$ of sequence $(a_n+b_n)_{n\in\mathbb{N}},$ we have \begin{gather*} \lim_{n\to\infty} (a_{\sigma(n)}+b_{\sigma(n)})\leq a^*+b^*+\epsilon. \end{gather*} Thus we see that $a^*+b^*+\epsilon$ is an upper bound of the set $C$ of subsequential limits of $(a_n+b_n)_{n\in\mathbb{N}}.$ Hence $c^*=\sup C\leq a^*+b^*+\epsilon.$ By the arbitrariness of $\epsilon>0,$ it follows that $c^*\leq a^*+b^*.$ $\Box$
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 $$ \limsup_{n\to\infty}a_n+\limsup_{n\to\infty}b_n\ge\limsup_{n\to\infty}(a_n+b_n)\tag2 $$
