Find two bounded sequences, $\{s_n\}_{n \in \mathbb N}$ and $\{t_n\}_{n \in \mathbb N}$, such that $\limsup(s_n+t_n))<\limsup(s_n)+\limsup(t_n)$ Find two bounded sequences, $\{s_n\}_{n \in \mathbb N}$ and $\{t_n\}_{n \in \mathbb N}$, such that $\limsup(s_n+t_n))<\lim(sup(s_n))+\limsup(t_n))$  
So I proved earlier that the inequality holds (well, replacing < with $\leq$).  The second part of the problem was just to find two sequences such that the strict inequality holds.  
My first thought was to use two sequences which don't contain their supremums, and add them.  But that turned out to be just equal.  Any suggestions on sequences which hold strict inequality?
 A: What do you get for these sequences?
$$\begin{align*}
s &= 0,1,0,1,0,1,0,1,\dots\\
t &= 1,0,1,0,1,0,1,0\dots
\end{align*}$$
Some intuition which might help to find examples like this.


*

*$\limsup s_n=S$ means that the sequence $s_n$ has many values "close to $S$" (and $S$ is the largest such value)

*$\limsup t_n=T$ means the sequence $t_n$ has many values "close to $T$" (and $T$ is the largest such value)

*If we have values close to $S$ and close to $T$ at the same $n$'s, then $\limsup(s_n+t_n)=S+T$.

*But if we get $\limsup s_n=S$ thanks to some positions and $\limsup t_n=T$ thanks to entirely different positions, then we get the desired counterexample. (In this case I took odd and even numbers, but there are many ways how this example can be modified.)



Of course, what I wrote is very vague, but if you look at definition of limit superior, this can be made more precise. 
There are several equivalent definitions of $\limsup x_n$, but whether you take the definition as the largest cluster point or as the largest subsequential limits or $\varepsilon$-definition, all of them are very close to the intuition described above. Perhaps the only of the commonly used definitions where connection with the handwavy description I've given above is not immediately clear is the definition as $\limsup\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty} (\sup\limits_{k\ge n} x_k$).
There are several posts on this site comparing various definitions of limit superior of a sequence. I have listed some of them here.
