设a0>0,an+1=arctanan(n=0,1,2,…). (Ⅰ)证明:存在; (Ⅱ)求

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问题 设a0>0,an+1=arctanan(n=0,1,2,…).
(Ⅰ)证明:存在;
(Ⅱ)求

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答案(Ⅰ)显然an>0(n=1,2,…), 又an+1=arctanan=arctanan-arctan0=[*]<an(0<ξ<an)得 {an)单调递减,所以[*]存在, 令[*]=A,an+1=arctanan两边求极限得A=arctanA, 解得A=0. (Ⅱ) [*]

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