Question

判断级数∞∑n=1 n^2/n!的敛散性

详细过程谢谢！

Answer 1

用比值法
|a(n+1)/an|
=[(n+1)^2/(n+1)!]/[n^2/n!]
=(n+1)^2/[n^2(n+1)]
=(n+1)/n^2
=1/n+1/n^2
->0 当n趋向∞

所以由比值判别法，此级数绝对收敛

追问

书面表达的话，好像有lim 能不能帮忙整理下，我其实不是太懂~不好意思啊

追答

都加lim就行
lim n->∞ |a(n+1)/an|
=lim n->∞ [(n+1)^2/(n+1)!]/[n^2/n!]
=lim n->∞ (n+1)^2/[n^2(n+1)]
=lim n->∞ (n+1)/n^2
=lim n->∞ 1/n+1/n^2
=0

Answer 2

这个是可以求和的
由级数e^x = 1 + x + x^2/2! + ...+ x^n/n! + ...
乘以x得xe^x = x + x^2 + x^3/2! + ...+x^(n+1)/n! + ....
求导数得
(x+1)e^x = 1 + 2x + 3x^2/2! + ... + (n+1)x^n/n! + ....
令x=1得
2e = 1 + 2 + 3/2! + ... + (n+1)/n! + ....
= ∑n^2/n!