Question

Chứng minh rằng

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}< 1\)

Answer 1

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..........+\frac{1}{2015^2}\)

\(\Leftrightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{2014.2025}\)

\(\Leftrightarrow B< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{2014.2015}\)

\(\Leftrightarrow B< 1-\frac{1}{2015}< 1\)

\(\Leftrightarrow B< 1\rightarrowđpcm\)

Answer 2

Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2014\cdot2015}\)

+ Xét : \(\frac{1}{1\cdot2}>\frac{1}{2^2}\)

\(\frac{1}{2\cdot3}>\frac{1}{3^2}\)

\(\frac{1}{3\cdot4}>\frac{1}{4^2}\)

...

\(\frac{1}{2015^2}< \frac{1}{2014\cdot2015}\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2014}-\frac{1}{2015}\)

\(A=1-\frac{1}{2015}< 1\)

\(\Rightarrow B< A< 1\left(đpcm\right)\)