I am looking for a sequence of sequences $(a_{n,i})_{n,i \in \mathbb N}$ with the following property:
$$\sum_{n \in \mathbb N}\limsup_{i\rightarrow \infty} a_{n,i} > \limsup_{i\rightarrow \infty} \sum_{n \in \mathbb N} a_{n,i}$$
I tried to think of primitive sequences for like 2 hours now but can't come up with one fulfilling this inequality. Can you help?
Let $s \colon \mathbb{N}\to \mathbb{N}$ a function that attains every value infinitely often, and set
$$a_{n,i} = \begin{cases} c_n &, n = s(i) \\ 0 &, n \neq s(i)\end{cases}$$
where $(c_n)$ is your favourite bounded strictly positive sequence. Then we have
$$\limsup_{i \to \infty} a_{n,i} = c_n$$
for every $n\in \mathbb{N}$, and
$$\sum_{n\in\mathbb{N}} \limsup_{i\to\infty} a_{n,i} = \sum_{n\in\mathbb{N}} c_n > \sup \{ c_n : n \in \mathbb{N}\} = \limsup_{i\to\infty} \sum_{n\in\mathbb{N}} a_{n,i}.$$
A relatively simple example for a function with the stated properties starts
$$0, 0,1, 0,1,2, 0,1,2,3, 0,1,2,3,4, 0,1,2,3,4,5,\dotsc.$$
We can give a closed form for that: for every $i\in\mathbb{N}$ there is exactly one $m\in \mathbb{N}$ with
$$\frac{m(m+1)}{2} \leqslant i < \frac{(m+1)(m+2)}{2},$$
namely
$$m = r(i) := \biggl\lfloor\frac{\sqrt{8i+1}-1}{2}\biggr\rfloor.$$
Then set
$$s(i) = i - \frac{r(i)\bigl(r(i)+1\bigr)}{2}.$$