Can we always establish whether an infinite series converges or diverges?

I'm not sure whether this is what you want or not. Consider the series $\displaystyle\sum_{n=1}^\infty a_n$, with$$a_n=\begin{cases}1&\text{ if }2^n-1\text{ is prime}\\0&\text{ otherwise.}\end{cases}$$It is not known whether it converges or not.

Just for fun, whatever foundational system $S$ you are working in, as long as $S$ can handle basic arithmetic, here is a series that $S$ (and hence you) cannot prove that it converges, and you hope $S$ never proves that it diverges! $ \def\nn{\mathbb{N}} $

Let $f : \nn \to \nn$ such that for each natural $n$, we have $f(n) = 1$ if there is a proof over $S$ of length at most $n$ of a contradiction, and $f(n) = 0$ otherwise.

Then $S$ cannot prove that $\sum_{n=0}^\infty f(n)$ converges (otherwise $S$ proves itself consistent, which contradicts Godel's incompleteness theorem).

And $S$ had better not prove that $\sum_{n=0}^\infty f(n)$ diverges (otherwise $S$ proves itself inconsistent).

A rather curious fact (if one has never seen the incompleteness phenomenon before) is that, if $S$ is consistent, then $S$ can actually prove that $f(0) = 0$ and $f(1) = 0$ and $f(1+1) = 0$ and so on, but cannot prove "$\forall n \in \nn\ ( f(n) = 0 )$".

One can improve this, via something equivalent to Rosser's trick, to obtain a series that $S$ (and hence you) cannot prove or disprove whether it converges!

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In general, any question of the form "Is it always possible to determine (prove or disprove) whether an object in collection $C$ satisfies property $P$?" is likely to have "no" as the answer if you can use $P$ on suitable members of $C$ to determine whether any given sentence over $S$ is provable or not.