Question

18．求證：$\frac{1}{1•\sqrt{1}}$+$\frac{1}{2•\sqrt{2}}$+$\frac{1}{3•\sqrt{3}}$+…+$\frac{1}{n•\sqrt{n}}$＜3．

Answer 1

分析 運(yùn)用放縮法證明．由$\frac{1}{\sqrt{{k}^{3}}}$＜$\frac{1}{\sqrt{（{k}^{2}-\frac{1}{4}）•k}}$=$\frac{\sqrt{2k-1}+\sqrt{2k+1}}{\sqrt{k}}$•（$\frac{1}{\sqrt{2k-1}}$-$\frac{1}{\sqrt{2k+1}}$）．可得$\frac{1}{\sqrt{{k}^{3}}}$＜2$\sqrt{2}$•（$\frac{1}{\sqrt{2k-1}}$-$\frac{1}{\sqrt{2k+1}}$）．再由裂項(xiàng)相消求和和不等式的性質(zhì)，即可得證．

解答 證明：$\frac{1}{\sqrt{{k}^{3}}}$＜$\frac{1}{\sqrt{（{k}^{2}-\frac{1}{4}）•k}}$=$\frac{2\sqrt{2}}{\sqrt{（2k-1）•2k•（2k+1）}}$
=$\frac{\sqrt{2k-1}+\sqrt{2k+1}}{\sqrt{k}}$•（$\frac{1}{\sqrt{2k-1}}$-$\frac{1}{\sqrt{2k+1}}$）．
其中（$\sqrt{2k-1}$+$\sqrt{2k+1}$）²=4k+2$\sqrt{4{k}^{2}-1}$＜4k+4k=8k，
則$\frac{1}{\sqrt{{k}^{3}}}$＜2$\sqrt{2}$•（$\frac{1}{\sqrt{2k-1}}$-$\frac{1}{\sqrt{2k+1}}$）．
即有$\frac{1}{1•\sqrt{1}}$+$\frac{1}{2•\sqrt{2}}$+$\frac{1}{3•\sqrt{3}}$+…+$\frac{1}{n•\sqrt{n}}$＜2$\sqrt{2}$[（1-$\frac{1}{\sqrt{3}}$）+（$\frac{1}{\sqrt{3}}$-$\frac{1}{\sqrt{5}}$）+…+（$\frac{1}{\sqrt{2n-1}}$-$\frac{1}{\sqrt{2n+1}}$）]
=2$\sqrt{2}$（1-$\frac{1}{\sqrt{2n+1}}$）＜2$\sqrt{2}$＜3．
故原不等式成立．

點(diǎn)評(píng) 本題考查不等式的證明，考查放縮法證明不等式的方法，考查推理能力和不等式的性質(zhì)，屬于難題．