# liminf of products of sequences

Dec. 17, 2013

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.