Ta có:
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{1}{5}.5=1\) (1)
\(\frac{1}{10}+\frac{1}{11}+...+\frac{1}{16}+\frac{1}{17}< \frac{1}{8}.8=1\) (2)
Cộng theo từng vế (1)và (2)
Ta được:
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}< 2\)
đầu tiên bạn tách tổng ra là hai 1 la từ 1/5 đến 1/9 còn lại tính rồi ss vs sao lớn nhất trong tổng đẫ tách ra đó
Ta có:
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{1}{5}.5=1\) (1)
\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{18}< \frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=\frac{1}{8}.8=1\) (2)
Cộng (1) và (2) ta có:
\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+.....+\frac{1}{18}< 1+1=2\)
\(\Rightarrow\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+....+\frac{1}{18}< 2\)
Vậy.........
Gọi \(A=\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}\)và \(B=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}\)
Ta có : \(\frac{1}{5}\)= \(\frac{1}{5}\); \(\frac{1}{6}< \frac{1}{5}\); \(\frac{1}{7}< \frac{1}{5}\); \(\frac{1}{8}< \frac{1}{5}\); \(\frac{1}{9}< \frac{1}{5}\)
\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+\frac{1}{5}\)
\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< \frac{1}{5}.5\)
\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}< 1\)
\(\Rightarrow A< 1\)(1)
Chứng minh tương tự với \(B\):
\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< \frac{1}{10}.8\)
\(\Rightarrow\)\(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{17}< 0.8\)
\(\Rightarrow B< 0.8\)(2)
Từ (1) và (2) \(\Rightarrow A+B< 1+0.8=2\)
\(\Rightarrow A+B< 1.8\)
\(\Rightarrow\)\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}...+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}< 1.8< 2\)
Vậy , ĐPCM
