(1!)^2/2!+(2!)^2/4!+…+(n!)^2/(2n)!+…是收敛,还是发散

(1!)^2/2!+(2!)^2/4!+…+(n!)^2/(2n)!+…是收敛,还是发散
(1!)^2/2!+(2!)^2/4!+…+(n!)^2/(2n)!+…是收敛,还是发散

(1!)^2/2!+(2!)^2/4!+…+(n!)^2/(2n)!+…是收敛,还是发散
收敛的,a(n) = (n!)^2/(2n)!
a(n+1) = (n+1)!^2/(2n+2)!
所以a(n+1) / a(n) = (n+1)^2 / (2n+2)(2n+1) = 1/4 < 1
又比值审敛法则可以知道这个级数收敛

用达兰贝尔判别法。xn=(n!)^2/(2n)!，则x(n+1)/xn={[(n+!)!]^2/[2(n+1)]!}/[(n!)^2/(2n)!]=(n+1)^2/[(2n+2)(2n+1)],则limx(n+1)/xn=1/4＜1，故原级数收敛。

判断级数条件收敛、绝对收敛还是发散，∑(n=1)(-1)^(n+1)*[2^(n^2)/n!]，求详细解答 级数写为求和（n＝1到无穷）(－1)^nan，则 a(n