Ta có:
1/2^2 < 1/1.2
1/3^2 < 1/2.3
...
1/2015^2 < 1/2014.2015
Suy ra: 1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 1/1.2 +1/2.3+...+1/2014.2015
1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 1-1/2+1/2-1/3+...+1/2014-1/2015
1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 1-1/2015
1/2^2 + 1/3^2 + 1/4^2+...+1/2015^2 < 2014/2015
Mình nghĩ đây là cách làm, bạn thử dựa vào làm xem nhé!
Ta có: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}=\frac{1}{2^2}+\left(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2015^2}\right)\)
Nhận xét: \(\frac{1}{2^2}=\frac{1}{4}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
......
\(\frac{1}{2015^2}< \frac{1}{2014.2015}\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2014.2015}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(\Rightarrow A< \frac{1}{4}+\left(\frac{1}{2}-\frac{1}{2015}\right)=\frac{1}{4}+\frac{1}{2}-\frac{1}{2015}=\frac{3}{4}-\frac{1}{2015}< \frac{3}{4}\)
Vậy A < 3/4
\(A=\frac{1}{4}+\left(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2015^2}\right)\)
\(A< \frac{1}{4}+\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2014.2015}\right)\)
\(=\frac{1}{4}+\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2014}-\frac{1}{2015}\right)\)
\(=\frac{1}{4}+\frac{1}{2}-\frac{1}{2015}=\frac{3}{4}-\frac{1}{2015}< \frac{3}{4}\)