Do the inverses of individual divergent subsequences of a divergent sequence always converge?
I'm thinking the inverse of a a divergent sequence needn't necessarily converge because take say the sequence $\{1/2,2,1/4,4,1/8,8,\ldots\}$; its inverse contains a divergent subsequence.
But if we split it into its two separate subsequences, we can invert the divergent subsequence and we have a pair of convergent sequences.
When can and can't a sequence be partitioned into subsequences which can be inverted to converge? And is this the essence of what's going on behind the scenes when we make an analytic continuation?
I imagine that by a process of repeated partitioning most divergent sequences (e.g. 95% of those found in a random sample taken from all divergent sequences mentioned on this site) can be treated like this and that potentially normal sequences such as the digits of $\pi$ which would perhaps cause the most difficulty. But is this known to be the case?
Every sequence in $\Bbb{R}$ has a monotone subsequence. If that subsequence does not converge, it is unbounded. Thus its multiplicative inverse will converge to zero. If on the other hand it does converge, you are done.
By deleting that monotone subsequence, you have a new sequence which must have a monotone subsequence. Continuing indefinitely you can extract monotone subsequences all of which will converge or their inverses will.
By tacking on some initial elements to the subsequences, you can guarantee that the entire original sequence is covered in this way, so I would say your intuition is correct here.