题目内容

4.已知数列{an}满足a1=$\frac{1}{2}$,an+1=$\frac{n{a}_{n}}{(n+1)(n{a}_{n}+1)}$(n∈N*).
(1)求数列{an}的通项公式;
(2)记Sn为数列{an}的前n项和,bn=(1-$\frac{{S}_{n}}{{S}_{n+1}}$)$\frac{1}{\sqrt{{S}_{n+1}}}$,求证:b1+b2+…+bn<2($\sqrt{2}$-1).

分析 (1)由已知得$\frac{1}{(n+1){a}_{n+1}}$=$\frac{n{a}_{n}+1}{n{a}_{n}}$=1+$\frac{1}{n{a}_{n}}$,从而{$\frac{1}{n{a}_{n}}$}是首项为2,公差为1的等差数列,由此能求出an=$\frac{1}{n(n+1)}$.
(2)由${a}_{n}=\frac{1}{n(n+1)}$=$\frac{1}{n}-\frac{1}{n+1}$,利用裂项求和法得Sn=$\frac{n}{n+1}$,从而bn=(1-$\frac{{S}_{n}}{{S}_{n+1}}$)$\frac{1}{\sqrt{{S}_{n+1}}}$<$2(\frac{1}{\sqrt{{S}_{n}}}-\frac{1}{\sqrt{{S}_{n+1}}})$,由此能证明b1+b2+…+bn<$2(\sqrt{2}-1)$.

解答 (1)解:∵数列{an}满足a1=$\frac{1}{2}$,an+1=$\frac{n{a}_{n}}{(n+1)(n{a}_{n}+1)}$(n∈N*),
∴$\frac{1}{{a}_{n+1}}$=$\frac{(n+1)(n{a}_{n}+1)}{n{a}_{n}}$,
∴$\frac{1}{(n+1){a}_{n+1}}$=$\frac{n{a}_{n}+1}{n{a}_{n}}$=1+$\frac{1}{n{a}_{n}}$,
又$\frac{1}{{a}_{1}}$=2,∴{$\frac{1}{n{a}_{n}}$}是首项为2,公差为1的等差数列,
∴$\frac{1}{n{a}_{n}}$=2+(n-1)×1=n+1,
∴an=$\frac{1}{n(n+1)}$;
(2)证明:∵${a}_{n}=\frac{1}{n(n+1)}$=$\frac{1}{n}-\frac{1}{n+1}$,
∴Sn=$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+…+\frac{1}{n}-\frac{1}{n+1}$=1-$\frac{1}{n+1}$=$\frac{n}{n+1}$,
∵$\frac{{S}_{n}}{{S}_{n+1}}=\frac{{n}^{2}+2n}{(n+1)^{2}}<1$,
∴bn=(1-$\frac{{S}_{n}}{{S}_{n+1}}$)$\frac{1}{\sqrt{{S}_{n+1}}}$=$(\frac{1}{{S}_{n}}-\frac{1}{{S}_{n+1}})•\frac{{S}_{n}}{\sqrt{{S}_{n+1}}}$=$(\frac{1}{\sqrt{{S}_{n}}}-\frac{1}{\sqrt{{S}_{n+1}}})(\frac{\sqrt{{S}_{n}}}{\sqrt{{S}_{n+1}}}+\frac{{S}_{n}}{{S}_{n+1}})$
<$2(\frac{1}{\sqrt{{S}_{n}}}-\frac{1}{\sqrt{{S}_{n+1}}})$,
则b1+b2+…+bn<$2(\frac{1}{\sqrt{{S}_{1}}}-\frac{1}{\sqrt{{S}_{2}}}+\frac{1}{\sqrt{{S}_{2}}}-\frac{1}{\sqrt{{S}_{3}}}+…+\frac{1}{\sqrt{{S}_{n}}}-\frac{1}{\sqrt{{S}_{n+1}}})$=$2(\frac{1}{\sqrt{{S}_{1}}}-\frac{1}{\sqrt{{S}_{n+1}}})=2(\sqrt{2}-\sqrt{\frac{n+2}{n+1}})$$<2(\sqrt{2}-1)$.

点评 本题考查数列的通项公式的求法,考查不等式的证明,解题时要认真审题,注意构造法、放缩法及裂项求和法的合理运用,是难度较大的题目.

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