Đặt:\(M=\frac{1}{2}\cdot\frac{3}{4}...\frac{9999}{10000}\)
\(N=\frac{2}{3}\cdot\frac{4}{5}...\frac{10000}{10001}\)
Dễ dàng nhận thấy: \(\frac{1}{2}
\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}......\frac{9999}{10000}<\frac{1}{100}\)
CMR
CMR
\(\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{9999}{10000}<\frac{1}{100}\)
Chứng tỏ rằng:C=\(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
Chứng tỏ rằng \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
Chứng minh rằng \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}< \frac{1}{100}\)
Chứng minh rằng:
\(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{9999}{10000}< \frac{1}{100}\)
chứng minh \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}<\frac{1}{100}\)
Bài 1 : Chứng minh
a) \(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
b) \(B=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
c) \(C=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.\frac{7}{8}...\frac{9999}{10000}< \frac{1}{100}\)
Cho A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9998}{9999}.\frac{10000}{10000}\)
So sánh A và 0,01
