I would agree with you assertion, namely:

If and that . To do this we note that in general where is the set of all convergent subsequential limits of and .

So, let denote the set of all subsequential limits of and the multiplication by of each subsequential limit of . We claim that .

We may assume (since otherwise this is easy) that . So, let then for some injection . Thus, we note that (notice that if a limit converges to a values so does every subsequential limit) and thus by the assumption that both of the limits converge we see that and thus is times a subsequential limit of and thus .

Conversely, if then and once again since we assumed both of these limits exists we see that and thus

Thus,

NOTE: It's late and I haven't checked over the above too carefully. I have been known to make light night stupid errors.