Question

证明n趋于无穷时(2n-1)!/(2n)!敛散性

Answer 1

设a_n = (2n-1)!/(2n)!,显然a_n > 0.

a_(n+1)/a_n = (2n+1)/(2n+2) < 1,故a_(n+1) < a_n,数列单调递减.

由其有下界0,故存在极限.

实际上ln((2n)!/(2n-1)!) = ln(1+1)+ln(1+1/3)+...+ln(1+1/(2n-1)) > 1/2+1/4+...+1/(2n)

n趋于∞时1/2+1/4+...+1/(2n)趋于∞,所以(2n)!/(2n-1)!趋于∞.

其倒数(2n-1)!/(2n)!趋于0.