Question

利用比值审敛法判定级数[∞ ∑ n=1] (n!)^2 / [(2n)!]的敛散性

Answer 1

an=(n!)^2/[(2n)!]
an+1/an=[(n+1)!]^2/[(2n+2)!]/(n!)^2/[(2n)!]
= [(n+1)!/n!]^2*[(2n)!/(2n+2)!]
=(n+1)^2/(2n+1)(2n+2)
lim(n→∞)an+1/an
=lim(n→∞) (n+1)^2/(2n+1)(2n+2)
=1/4 <1
则该正项级数收敛

追问

= [(n+1)!/n!]^2*[(2n)!/(2n+2)!]
上式怎么等于下式的?
=(n+1)^2/(2n+1)(2n+2)

追答

[(n+1)!/n!
=[(n+1)*n*(n-1)*(n-2)……*1]/[n*(n-1)*(n-2)……*1]
=(n+1) 写出来就发现后面的都消了
(2n)!/(2n+2)!
=[(2n)*(2n-1)*(2n-2)……*1]/[(2n+2)*(2n+1)*2n*(2n-1)……*1]
=1/(2n+1)(2n+2) 写出来发现分子都消去了