Chứng minh rằng : 1/65 < 1/5^3 + 1/6^3 + 1/7^3 + ... + 1/2004^3 <1/40
\(choA=\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+...+\dfrac{1}{n^2}+...+\dfrac{1}{2004^2}\)
\(chứngminh\dfrac{1}{65}< A< \dfrac{1}{4}\)
chứng minh rằng \(\frac{1}{65}\)<\(\frac{1}{5^3}+\frac{1}{6^3}+...+\frac{1}{2004^3}\)<\(\frac{1}{40}\)
chung minh : 1/5^3+1/6^3+1/7^3+........+1/2004^3 <1/40
chung minh rang 1/5^3+1/6^3+1/7^3+..........+1/2004^3<1/40
chung minh : 1/5^3+1/6^3+1/7^3+.........+1/2004^3 < 1/40
1/5^3+1/6^3+...+1/2004^3<1/40
cho \(A=\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+....+\frac{1}{2004^3}\) và \(B=\frac{1}{65}\)
So sánh A và B
chứng minh rằng
\(\frac{1}{5^3}+\frac{1}{6^3}+\frac{1}{7^3}+...+\frac{1}{2004^3}\)<\(\frac{1}{40}\)