## Calculus: Early Transcendentals 8th Edition

We can write $(2n + 1)!$ as $(2n + 1)(2n)(2n - 1)!$. Hence $$a_n = \frac{(2n-1)!}{(2n + 1)!} = \frac{(2n - 1)!}{(2n + 1)(2n)(2n - 1)!} = \frac{1}{4n^2 + 2n}.$$ Hence the sequence converges to 0.