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.


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

Then, \[\liminf a_nb_n = AB.\]


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.


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.