數學知識交流 － 求值

(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)
.
.
.
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?

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

the series is not converges , (n^3-1)/(n^3+1) -> 1 =/= 0 as n->inf.