數學知識交流 － 求值
回答 (3)
2012-01-20 11:53 pm
✔ 最佳答案
(2^3-1)/(2^3+1) + (3^3-1)/(3^3+1) + (4^3-1)/(4^3+1) +... a1=((1+1)^3-1)/((1+1)^3+1)
a2=((2+1)^3-1)/((2+1)^3+1)
a3=((3+1)^3-1)/((3+1)^3+1)
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an=((n+1)^3-1)/((n+1)^3+1)
=1-(2/(n+1)^3+1)
當n趨向無限，-(2/(n+1)^3+1)趨向0
所以an=1
原式=1+1+1+1+...
這式發散(+無限)
PS1﹕做這些題目是這樣的
PS2﹕沒有暈倒嗎?
參考: MY BRAIN
2012-01-20 8:28 pm
7/9+26/28+63/65+.......=7/9+13/14+63/65+........=21913/8190+.......=2 5533/8140+......
2012-01-20 12:30:15 補充：
less than 1, but it's impossable to be 0
2012-01-20 12:30:15 補充：
less than 1, but it's impossable to be 0
2012-01-20 7:31 am
the series is not converges , (n^3-1)/(n^3+1) -> 1 =/= 0 as n->inf.
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https://hk.answers.yahoo.com/question/index?qid=20120119000051KK00868