Re: [歛散] 請教一題級數審歛

[serendipper (Ser)]

※ 引述《jjerry8888 (夏蘭行德楓)》之銘言：
: ∞
: Σ  n^2e^(-n^3)
: n=1
: 此題是要用哪一種方法判斷歛散性啊?

 ∞                              ∞
∫ 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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