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 $a_n$ and $b_n$ be two sequences such that $\liminf a_n = A$ and $\lim b_n = B, 0<B<\infty$.

Then, $\liminf a_nb_n = AB.$

## Proof

Since $B>0$ it must be the case for all sufficiently large $n$ that $2B>b_n>B/2>0$. 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 $a_{n_k}$ converges to a limit $\alpha$ if and only if the corresponding subsequence $a_{n_k}b_{n_k}$ converges to $\alpha B$. If we can establish this fact, then, since $B>0$, the infimum of the subsequential limits of $a_nb_n$ will be $B$ times the infimum of the subsequential limits of $a_n$.

When $\alpha$ is finite, the required fact is a direct consequence of the usual theorems on limits of products and quotients.

For $\alpha$ infinite we use the fact that $b_n$ has positive lower and upper bounds, so $a_{n_k}b_{n_k}$ will be unbounded above or below if and only if $a_{n_k}$ is unbounded in the same way.

## Notes

The condition $B>0$ is essential in the above theorem. Consider the following example: $a_n = 2,2,2,4,2,6,2,8,2,10,\ldots$. $b_n = -1,-1/2,-1/3,-1/4,\ldots$. Then $\liminf a_n = 2$, $\lim b_n = 0$, but $\liminf a_nb_n = -1$.

We can get by with $B \ge 0$ if we are willing to assume that $a_n$ is bounded. But not having to check boundedness when $B>0$ is a big help.