We have the result that $\limsup_{n\rightarrow \infty} (a_n + b_n) \leq \limsup_{n\rightarrow \infty}(a_n) + \limsup_{n\rightarrow \infty}(b_n)$ if all sums exist.
By induction, we can extend this to any finite sum of terms.
Does this mean we can take the limit and say $\limsup_{n\rightarrow \infty} (\sum_{k=1}^{\infty} a_n^k) \leq \sum_{k=0}^\infty \limsup_{n\rightarrow \infty} a_n^k$? (if all sums exist)
Intuitively I feel like this is true, but I don't know how to pass the limit through the $\limsup$ to prove this.
Best Answer
So indeed the answer is "no." Reiterating the counter-example from my comment, consider: $$ a_n^k = \left\{\begin{array}{cc} 1 & \mbox{ if $n=k$} \\ 0 & \mbox{ else} \end{array}\right.$$ The proof for the finite case relies on the fact that for all $\epsilon>0$, eventually all (finite) terms in the sum are no more than $\epsilon$ beyond their $\limsup$s. That is not true for the infinite sum case.
However, an application of Fatou's lemma ensures that if $a_n^k\geq 0$ for all $n,k$ then $$ \liminf_{n\rightarrow\infty} \sum_{k=1}^{\infty} a_n^k \geq \sum_{k=1}^{\infty} \liminf_{n\rightarrow\infty} a_n^k$$ although you can also prove this directly (without Fatou's lemma).