例题1-双阶乘

题目:

$$求证: \dfrac{1}{2\sqrt{n}}<\dfrac{(2n-1)!!}{(2n)!!}<\dfrac{1}{\sqrt{2n-1}}, (n \geq 2)$$

解答:

$$令a_n = \frac{(2n-1)!!}{(2n)!!} = \frac{1}{2} \times \frac{3}{4} \times \cdots \times \frac{2n-1}{2n} \\ 令b_n = \frac{1}{(2n+1)a_n} = \frac{2}{3} \times \frac{4}{7} \times \cdots \times \frac{2n}{2n+1} \\ 因为\frac{2n-1}{2n} < \frac{2n}{2n+1}, 所以 a_n < b_n \\ \implies a_n^2 < a_n b_n = \frac{1}{2n+1} \implies a_n < \frac{1}{\sqrt{2n+1}} < \frac{1}{\sqrt{2n-1}} \\ 故不等式右侧成立 \\ \\ \\ n=1时, \frac{1}{2\sqrt 1} = \frac{1!!}{2!!} = \frac{1}{2} 成立 \\ 假设n=k时, \frac{1}{2 \sqrt k} < \frac{(2k-1)!!}{(2k)!!} \\ 则当n=k+1时, \frac{(2k+1)!!}{(2k+2)!!} = \frac{2k+1}{2n+2} \frac{(2k-1)!!}{(2k)!!} > \frac{2k+1}{2n+2} \cdot \frac{1}{2 \sqrt k} \\ 只需证 \frac{2k+1}{2n+2} \cdot \frac{1}{2 \sqrt k} > \frac{1}{2\sqrt{k+1}} \iff \frac{(2k+1)^2}{4(k+1)^2} > \frac{k}{k+1} \\ \iff \frac{4k+3}{4(k+1)^2} < \frac{1}{k+1} \iff \frac{4k^2+7k+3}{4k^2+8k+4} < 1 \\ 因为k>0, 故原式成立, 由数学归纳法得: \frac{1}{2\sqrt n} < \frac{(2n-1)!!}{(2n)!!} \\ 故不等式左侧成立 \\$$