学数答题160913-数列求和有界

题160913（14分）数列$\left\{ {{a}_{n}} \right\}$满足：${{a}_{1}}=1$，${{a}_{n+1}}={{a}_{n}}+2n+1$，求证：$\sum\limits_{i=1}^{n}{\dfrac{1}{{{a}_{i}}}}<\dfrac{5}{3}$．

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题目来源：2013年中国科学技术大学入学考试

证明：由${{a}_{n+1}}={{a}_{n}}+2n+1$，得${{a}_{n+1}}-{{a}_{n}}=2n+1$，

累加得${{a}_{n}}=\left( {{a}_{n}}-{{a}_{n-1}} \right)+\left( {{a}_{n-1}}-{{a}_{n-2}} \right)+\cdots +\left( {{a}_{2}}-{{a}_{1}} \right)+{{a}_{1}}$，

即${{a}_{n}}=\left[ 2\times \left( n-1 \right)+1 \right]+\left[ 2\times \left( n-2 \right)+1 \right]+\cdots +\left[ 2\times 1+1 \right]+1$，

得${{a}_{n}}=1+2\times \dfrac{n\left( n-1 \right)}{2}+\left( n-1 \right)={{n}^{2}}$，

因此，$\sum\limits_{i=1}^{n}{\dfrac{1}{{{a}_{i}}}=}\sum\limits_{i=1}^{n}{\dfrac{1}{{{i}^{2}}}}$

$<1+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+\left[ \dfrac{1}{5\times 6}+\dfrac{1}{6\times 7}+\cdots +\dfrac{1}{\left( n-1 \right)n} \right]$

$=1+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{5}-\dfrac{1}{n}$

$=1+\dfrac{2389}{3600}-\dfrac{1}{n}$$<\dfrac{5}{3}$．