Question

双阶乘：lim ((2n-1)!/(2n)!) n趋近无穷大

Answer 1

0 < ((2n-1)!/(2n)!) ^2

=[1*3*3*5*5*...*(2n-3)*(2n-1)*(2n-1)] *【(2n+1)/(2n+1)】/[2*2*4*4*...*2n*2n]

=[1*3/2*2]*[3*5/4*4]*[5*7/6*6]*...*[(n-1)(n+1)/n^2] /(2n+1)

∵ [(k-1)(k+1)/k^2 < 1 ∴

< 1/(2n+1) -->0 (n->∞时)

∴ lim(n->∞) ((2n-1)!/(2n)!) = 0