Convergence, divergence and existence

Question

Is true that every limit can only converge, diverge or(exclusive) not exist?

Can I demonstrate that it doesn't exist after I proved it doesn't converge neither diverge?

I've never seen this, but it makes some sort of sense to me. If a real isn't positive nor negative, it must be zero... But with limits.

Answer 1

Divergence means the limit doesn't exist. "Divergence to $\infty$" is a special case of divergence, and we sometimes say that the limit exists in those cases, but strictly speaking it doesn't (unless we're working in the extended reals, which as far as I can tell is mostly done just to indulge in this specific abuse of terminology (yes, I know there are legitimate reasons to use them, I was was only being half-serious)).

So yes, a sequence can only converge or diverge, because either there is a limit, or there isn't. If you like the four categories "converge", "diverge to $\infty$", "diverge to $-\infty$" and "diverge, but not to any infinity", then yes, those four categories cover everything, because the last one specifically covers everything not in the first three, by definition.

Answer 2

In the following, I'll use $n$ for $n\to +\infty$ and $x$ for $x\to 0$.

Converge means it has a limit and this limit is finite.

• For instance $\frac 1n$ converges to $0$

Diverge means does not converge.

There are multiple forms of divergence

• The limit is infinite $\frac 1{x^2}\to+\infty$, so it has a limit in the extended sense.

• The limit is infinite but with undefined sign $\frac 1x\to\pm\infty$ whether $x\to 0^+$ or $x\to 0^-$, now even in the extended sense, it has no limit or we say it doesn't exists.

• There are multiple adherence points $(-1)^n\frac {n+1}{n+2}$ has two limit points $1$ and $-1$ (better said two convergent subsequences), even worst is the case $\sin(\frac 1x)$ where whole $[-1,1]$ is adherent.

• There are some wild values for instance $\dfrac 1{n\sin(\frac{n\pi}8)}$, it "mostly" goes to $0$ but from time to time there are infinite values. Though they don't need to be infinite, take $f(n)=\frac 1n$ and $f(n^2)=\text{random}$, it doesn't converge either.

These are only some common examples of divergence, it is not exhaustive, as soon as the values are not all concentrated around one single finite value, there is divergence.