作者serendipper (Ser)
看板trans_math
标题Re: [歛散] 请教一题级数审歛
时间Thu Jun 23 11:21:11 2011
※ 引述《jjerry8888 (夏兰行德枫)》之铭言：
: ∞
: Σ n^2e^(-n^3)
: n=1
: 此题是要用哪一种方法判断歛散性啊?
∞ ∞
∫ n^2 e^(-n^3) dx = -e^(n^3)/3 ] = 1/3e
1 1
By the Integral Test, it converges.
lim [a_(n+1) / a_n]
n→∞
= lim (n+1)^2 / e^[(n+1)^3] * e^(n^3) / n^2
= lim (n+1)^2/n^2 * e^[n^3-(n+1)^3]
= 1．lim e^[-3n^2-3n-1] = 1．0 < 1
By the Ratio Test, it converges.
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1F：推 jjerry8888:感谢 163.25.118.163 06/23 11:39