Question

15．$\underset{lim}{n→∞}$（$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$）=0．

Answer 1

分析 由題意知$\frac{2n-1}{2n}$＜$\frac{2n}{2n+1}$，從而可得0＜$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$＜$\frac{1}{\sqrt{2n+1}}$，從而求極限．

解答 解：∵$\frac{2n-1}{2n}$＜$\frac{2n}{2n+1}$，
∴（$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$）²=$\frac{1}{2}$×$\frac{1}{2}$×$\frac{3}{4}$×$\frac{3}{4}$×…×$\frac{2n-1}{2n}$×$\frac{2n-1}{2n}$
＜$\frac{1}{2}$×$\frac{2}{3}$×$\frac{3}{4}$×$\frac{4}{5}$×…×$\frac{2n-1}{2n}$×$\frac{2n}{2n+1}$=$\frac{1}{2n+1}$，
∴0＜$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$＜$\frac{1}{\sqrt{2n+1}}$，
又∵$\underset{lim}{n→∞}$0=0，$\underset{lim}{n→∞}$$\frac{1}{\sqrt{2n+1}}$=0，
∴$\underset{lim}{n→∞}$（$\frac{1}{2}×\frac{3}{4}×\frac{5}{6}…\frac{2n-1}{2n}$）=0，
故答案為：0．

點(diǎn)評 本題考查了極限的求法．