Unusual topologies in $\mathbb C$

Question

Given complex-valued sequences, there are several conditions you can put on those sequences, which all make the set of sequences fulfilling those conditions into vector spaces:

1. You can demand nothing
2. You can demand that almost all $a_n=0$.
3. You can demand that the sequence converges to zero.
4. You can demand that the sequence converges.
5. You can demand that the series $\sum_n a_n$ converges.
6. You can demand that the series $\sum_n a_n$ converges absolutely.
7. You can demand that $\sum_n a_n x^n$ is an entire function.

Of course this list is far from exhaustive.

Now it occurred to me that case 1 is equivalent to demanding that the sequence converges to $0$ in the trivial topology (in which, of course, every sequence converges to every value). The trivial topology is of course compatible with the field structure of $\mathbb C$ (because all functions into a trivial topology are continuous).

Also case 2 is equivalent to demanding that the sequence converges to $0$, but this time in the discrete topology. Again, that topology is compatible with the field structure of $\mathbb C$ as any function whose domain is the discrete topology is continuous.

And of course case 3 is quite literally demanding convergence to $0$, in the standard topology of $\mathbb C$.

Therefore I wonder:

Can some, or even all, of the other conditions also be expressed as the condition that the sequence converges in an appropriate topology that is compatible with the field structure of $\mathbb C$?

And if the answer is "some, but not all", is there an easy rule to identify those where it is possible?

Answer 1

Convergence (in the usual topology) of a series can never be equivalent to convergence of the sequence in some topology. This is simply because the seqeunce $$1,\frac14,\frac19,\frac1{16},\frac1{25},\ldots $$ leads to the convergent series $\sum\frac1{n^2}$ and $$1,\underbrace{\frac14,\frac14}_2,\underbrace{\frac19,\frac19,\frac19}_3,\underbrace{\frac1{16},\frac1{16},\frac1{16},\frac1{16}}_4,\underbrace{\ldots\qquad\vphantom{\frac11}}_\ldots $$ leads to the divergent $\sum\frac 1n$.