Products of ratios of consecutive integers

Take the product of the numbers (n/(n + 1))(is an element of n) for 1 <= n < N, where each is an element of(n) is +/- 1. Express the product as a/b in lowest terms. Evidently the minimal possible value for a over all choices for is an element of(n) is 1; just take each is an element of(n) = 1, or each is an element of(n) = 0. Denote the maximal possible value of a by A( N). It is known from work of Nicolas and Langevin that ( log 4 + o( 1)) N <= log A( N) <= (2/ 3 + o(1)) N log N. Using the Rosse-Iwaniec sieve, we improve the lower bound to the same order of magnitude as the upper bound.