Re: [歛散] 請教一題級數審歛

[rygb (再生)]

※ 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
------- = an   an always > 0 and an+1 < an
 (n^3)
e


Use Integral Test 

 1       1      3     1   -(n^3)  |n=0          1
-- ∫ ------ d n   = --- e        |         = ----   so its converge
 3      n^3           3           |n=∞         3
       e


Use Limit Comparison Test

                         an
we know that if   lim   ---- = 0 and Σbn converge , then Σan converge
                 n->∞   bn
          1
Σbn = Σ---
           2
          n



          an
 lim    -----  = 0       and  Σbn is p-series p >2 so  Σbn converge
n->∞     bn


then Σan converge


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◆ From: 114.34.122.244
※ 編輯: rygb            來自: 114.34.122.244       (06/23 11:24)
推 jjerry8888:感謝囉!!!