Produce sequences \(\{a_n\},\,\{b_n\}\) of positive real numbers such that \[\liminf_{n\to\infty}(a_nb_n)>\left(\liminf_{n\to\infty} a_n\right) \left(\liminf_{n\to\infty} b_n\right).\]
If \(\{a_n\},\,\{b_n\}\) are sequences of positive real numbers and \(\{a_n\}\) converges, prove that \[\liminf_{n\to\infty}(a_nb_n)=\left(\lim_{n\to\infty}a_n\right)\left(\liminf_{n\to\infty}b_n\right).\]
Solution
1. (Part 1) Take \(a_n = 2 + (-1)^n\) and \(b_n = 2 - (-1)^n\).
\(a_n = 3, 1, 3, 1, \ldots\) and \(b_n = 1, 3, 1, 3, \ldots\), both positive.
2. (Part 2) Let \(a_n \to a\) with \(a_n, b_n > 0\).Then \(\liminf(a_n b_n) = a \liminf b_n\).
\(a > 0\) since \(a_n > 0\) for all \(n\) (the limit of positive numbers is \(\ge 0\); if \(a = 0\) then \(0 \le \liminf a_n b_n\), and for any subsequence \(a_n b_n\) along it tends to \(0\) because \(a_n \to 0\) and \(b_n\) is bounded below by \(0\), so \(\liminf a_n b_n = 0\)). For \(a > 0\): since \(a_n \to a\), for any \(\eps\) with \(0 < \eps < a/2\) there is \(N\) with \(a - \eps \le a_n \le a + \eps\) for \(n \ge N\).
3. Q.E.D.