I needed a result about the liminf of a product of two sequences that I could not find in the references I had immediately at hand. Noting it down for future reference.
Theorem
Let an and bn be two sequences such that liminf an = A and lim bn = B, 0 < B < ∞.
Then, liminf anbn = AB.
Proof
Since B > 0 it must be the case for all sufficiently large n that 2B > bn > B/2 > 0. Since all the properties asserted in the theorem depend only on the tail behaviour of the sequences there is no harm in assuming that this inequality holds for all n.
We make use of the definition of liminf as the infimum of subsequential limits (Rudin, Principles of Mathematical Analysis, 3rd edition, Definition 3.16).
The theorem will be proved if we can show that any subsequence ank converges to a limit α if and only if the corresponding subsequence ankbnk converges to αB. If we can establish this fact, then, since B > 0, the infimum of the subsequential limits of anbn will be B times the infimum of the subsequential limits of an.
When α is finite, the required fact is a direct consequence of the usual theorems on limits of products and quotients.
For α infinite we use the fact that bn has positive lower and upper bounds, so ankbnk will be unbounded above or below if and only if ank is unbounded in the same way.
Notes
The condition B > 0 is essential in the above theorem. Consider the following example: an = 2, 2, 2, 4, 2, 6, 2, 8, 2, 10, …. bn = − 1, − 1/2, − 1/3, − 1/4, …. Then liminf an = 2, lim bn = 0, but liminf anbn = − 1.
We can get by with B ≥ 0 if we are willing to assume that an is bounded. But not having to check boundedness when B > 0 is a big help.