题目
设__ _(n)=(int )_(0)^dfrac (pi {4)}(tan )^nxdx geqslant 1.-|||-(1)UND明数UND(a)_(n)ykparallel UND数;-|||-(2)证明 _(n)+(a)_(n-2)=dfrac (1)(n-1) , gt 2 ;-|||-(3)证明 dfrac (1)(2(n+1))lt (a)_(n)lt dfrac (1)(2(n-1)).
题目解答
答案
解析
步骤 1：证明数列{an}收敛
当 $0\lt x\lt \dfrac {\pi }{4}$ 时, $0\lt \tan x\lt 1$ .则 .$0\lt {\tan }^{n+1}\quad x\lt {\tan }^{n}x$ - 因此 0步骤 2：证明 ${a}_{n}+{a}_{n-2}=\dfrac {1}{n-1}$ , $n\gt 2$
${a}_{n}+{a}_{n-2}={\int }_{0}^{\dfrac {n}{4}}({\tan }^{n}x+{\tan }^{n-2}x)dx$ . .$={\int }_{0}^{\dfrac {\pi }{4}}{\tan }^{m-2}x({\tan }^{2}x+1)dx$ .$=\dfrac {1}{n-1}{\tan }^{n-1}x{\int }_{0}^{\dfrac {\pi }{4}}=\dfrac {1}{n-1}$
步骤 3：证明 $\dfrac {1}{2(n+1)}\lt {a}_{n}\lt \dfrac {1}{2(n-1)}$
由于数列{an }单调减少,有 .$2{a}_{n}\lt {a}_{n}+{a}_{n-2}=\dfrac {1}{n-1}$ 因此 ${a}_{n}\lt \dfrac {1}{2(n-1)}$ 同理 .$2{a}_{n}\gt {a}_{n}+{a}_{n+2}=\dfrac {1}{n+1}$ ,因此 .$\dfrac {1}{2(n+1)}\lt {a}_{n}\lt \dfrac {1}{2(n-1)}$
当 $0\lt x\lt \dfrac {\pi }{4}$ 时, $0\lt \tan x\lt 1$ .则 .$0\lt {\tan }^{n+1}\quad x\lt {\tan }^{n}x$ - 因此 0
${a}_{n}+{a}_{n-2}={\int }_{0}^{\dfrac {n}{4}}({\tan }^{n}x+{\tan }^{n-2}x)dx$ . .$={\int }_{0}^{\dfrac {\pi }{4}}{\tan }^{m-2}x({\tan }^{2}x+1)dx$ .$=\dfrac {1}{n-1}{\tan }^{n-1}x{\int }_{0}^{\dfrac {\pi }{4}}=\dfrac {1}{n-1}$
步骤 3：证明 $\dfrac {1}{2(n+1)}\lt {a}_{n}\lt \dfrac {1}{2(n-1)}$
由于数列{an }单调减少,有 .$2{a}_{n}\lt {a}_{n}+{a}_{n-2}=\dfrac {1}{n-1}$ 因此 ${a}_{n}\lt \dfrac {1}{2(n-1)}$ 同理 .$2{a}_{n}\gt {a}_{n}+{a}_{n+2}=\dfrac {1}{n+1}$ ,因此 .$\dfrac {1}{2(n+1)}\lt {a}_{n}\lt \dfrac {1}{2(n-1)}$