paciii, f-ooo. I think paciii's answer 0.25, is most likey the rite answer
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作者: sigmaxi ® 02/18/2009, 16:54:01 作者简介 电邮作者 修改

I'll verify with my coleague -- do high school students in USA learn calculus? below by paciii details: sum= intg1 + intg2 intg1 = integral ( (2x)^2 / (1+ (2x)^3) ), x=1, ... n intg2 = integral ( (2x+1)^2/(1+(2x+1)^3) ), x=0,...n solve it, we get: (变上限定积分） intg1 = -1/6ln[1+(2x)^3], x:1,n intg2 = 1/6ln[1+(2x+1)^3], x: 0, n together: sum = 1/6ln9 - 1/6ln2 + ln[(1+(2n+1)^3)/(1+(2n)^3] = 1/6ln4.5 + 0 ; when n approaches to inifinity. for question

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paciii you are truely genius. (无内容) --- sigmaxi ( 02/18/2009, 16:59:37 ) ( 0 bytes ) +1 过奖过奖。 (无内容) --- paciii ( 02/18/2009, 17:17:07 ) ( 0 bytes ) +1 Math PhD? (无内容) --- sigmaxi ( 02/18/2009, 17:21:49 ) ( 0 bytes ) +1 No. Msc computer science. (无内容) --- paciii ( 02/18/2009, 17:25:04 ) ( 0 bytes ) +1 that is amazing. (无内容) --- sigmaxi ( 02/18/2009, 17:37:08 ) ( 0 bytes ) As I said, the integral representation of the sum is not exact for the leading --- f-000 ( 02/18/2009, 17:08:26 ) ( 131 bytes ) +1 It converges for sure, but must be less than the value of the first term --- sigmaxi ( 02/18/2009, 17:13:00 ) ( 12 bytes ) +1 I misread paciii's procedure. It is convergent when the limit is taken after --- f-000 ( 02/18/2009, 18:09:15 ) ( 24 bytes ) +1 0.239561 is the right answer. (无内容) --- f-000 ( 02/18/2009, 18:19:39 ) ( 0 bytes ) I think there is a problem --- Nono ( 02/18/2009, 19:52:26 ) ( 224 bytes ) +1 It was treated as two finite sums, and represent the sums as integrals, then --- f-000 ( 02/18/2009, 20:12:03 ) ( 189 bytes ) +1 You will have problem getting a finit limit as n goes to infinity --- Nono ( 02/18/2009, 20:38:37 ) ( 19 bytes )