其實可以更簡單一點,
Let {an}={1,1,-2,1/2^(1/3),1/2^(1/3),-2/2^(1/3),1/3^(1/3),1/3^(1/3),-2/3^(1/3),
Hence sum an = 0
and sum sin(an)=sum (2sin(1/n^(1/3)) - sin(2/n^(1/3)))
Note that 2sin(x)-sin(2x) = x^3 + O(x^5).
Thus, sum sin(an)= sum (1/n + O(1/n^(5/3))
= +00.
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 118.165.201.183
※ 文章網址: https://www.ptt.cc/bbs/Math/M.1433263887.A.956.html
※ 編輯: motivic (118.165.188.231), 06/03/2015 10:45:02
※ 編輯: motivic (118.161.25.38), 06/03/2015 14:56:27