题目:
求证:2n1<(2n)!!(2n−1)!!<2n−11,(n≥2)
解答:
令an=(2n)!!(2n−1)!!=21×43×⋯×2n2n−1令bn=(2n+1)an1=32×74×⋯×2n+12n因为2n2n−1<2n+12n,所以an<bn⟹an2<anbn=2n+11⟹an<2n+11<2n−11故不等式右侧成立n=1时,211=2!!1!!=21成立假设n=k时,2k1<(2k)!!(2k−1)!!则当n=k+1时,(2k+2)!!(2k+1)!!=2n+22k+1(2k)!!(2k−1)!!> 2n+22k+1⋅2k1只需证2n+22k+1⋅2k1>2k+11⟺4(k+1)2(2k+1)2>k+1k⟺4(k+1)24k+3<k+11⟺4k2+8k+44k2+7k+3<1因为k>0,故原式成立,由数学归纳法得:2n1<(2n)!!(2n−1)!!故不等式左侧成立