题目内容
17.证明:$\frac{1}{2n+1}$<$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2n-1}{2n}$<$\frac{1}{\sqrt{2n+1}}$(其中n∈N*).分析 利用数学归纳法证明即可.
解答 证明:下面用数学归纳法来证明:
(1)先证明:$\frac{1}{2n+1}$<$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2n-1}{2n}$;
①当n=1时,命题显然成立;
②假设当n=k(k≥2)时,有$\frac{1}{2k+1}$<$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2k-1}{2k}$,
则$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2k-1}{2k}$•$\frac{2k+1}{2(k+1)}$>$\frac{1}{2k+1}$•$\frac{2k+1}{2(k+1)}$=$\frac{1}{2(k+1)}$,
即当n=k+1时,命题也成立;
由①、②可知$\frac{1}{2n+1}$<$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2n-1}{2n}$;
(2)再证明:$\frac{1}{2n+1}$<$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2n-1}{2n}$<$\frac{1}{\sqrt{2n+1}}$;
①当n=1时,命题显然成立;
②假设当n=k(k≥2)时,有$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2k-1}{2k}$<$\frac{1}{\sqrt{2k+1}}$,
则$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2k-1}{2k}$•$\frac{2k+1}{2(k+1)}$<$\frac{1}{\sqrt{2k+1}}$•$\frac{2k+1}{2(k+1)}$
=$\frac{\sqrt{2k+1}}{2k+2}$
<$\frac{\sqrt{2k+1}}{2k+1}$
=$\frac{1}{\sqrt{2k+1}}$,
即当n=k+1时,命题也成立;
由①、②可知$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2n-1}{2n}$<$\frac{1}{\sqrt{2n+1}}$;
综上所述,$\frac{1}{2n+1}$<$\frac{1}{2}$•$\frac{3}{4}$•$\frac{5}{6}$•…•$\frac{2n-1}{2n}$<$\frac{1}{\sqrt{2n+1}}$.
点评 本题考查不等式的证明,利用数学归纳法是解决本题的关键,注意解题方法的积累,属于中档题.
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