For the sequence $$S_n=((-1)^j\times j); \forall j \in \mathbb{N}$$
I am having hard time understanding what
Limit of the sequence does not exist but $$\limsup_{n\rightarrow\infty}(S_n)$$ exist and =$\infty$ and $$\liminf_{n\rightarrow\infty}(S_n)$$ exists and =$-\infty$ mean.
Added: My only concern here is in case of limit, whenever limit is equal to infinity, we say limit does not exist. But
Why does Limit sup =infinity mean limit exists? Why is there such difference?
Best Answer
The $\limsup$ is the largest real number, or $\pm\infty$, which is a limit of a subsequence of $S_n$, whereas $\liminf$ is the smallest real number (or $\pm\infty$) which is a limit of a subsequence of $S_n$.
It follows that $\lim a_n$ exists (in the broad sense) if and only if $\liminf a_n=\limsup a_n$.
The existence of $\liminf,\limsup$ follows from the completeness of $\mathbb R$.
If $S_n$ is a sequence such that $\limsup S_n=\infty$ then there is a subsequence, $S_{n_k}$ which is strictly increasing. Therefore its limit is $\infty$.
To the edit:
One cannot treat $\infty$ as a real number. It's good when a sequence has a limit within the real numbers, but sometimes it is sufficient that it is convergent, i.e. satisfies some definition.
To say that the limit of a sequence is $\infty$ is to say that although the sequence does not tend to a real number, it behaves "nice enough". This is in contrast to sequences which have no limit at all and just jump around between numbers.
This is why we sometimes say that it converges in a broad sense when it has the limit $\pm\infty$. Sometimes, when context is clear enough, we may omit the "broad sense" part too.