Ta có :
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}\)
\(=1-\frac{1}{2013}< 1\)( đpcm )
\(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
....
\(\frac{1}{2013^2}< \frac{1}{2012.2013}=\frac{1}{2012}-\frac{1}{2013}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2012}-\frac{1}{2013}=1-\frac{1}{2013}< 1\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2013^2}\)
Ta thấy \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{2013^2}\)\(< \frac{1}{2012.2013}\)
\(\Rightarrow A< B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2012.2013}\)
MÀ \(B=1-\) \(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2013}\)
\(=1-\frac{1}{2013}< 1\)
Mà A<B nên A<1(t/c)(đpcm)
ta thấy \(\frac{1}{2^2}\)<\(\frac{1}{1.2}\);\(\frac{1}{3^2}\)<\(\frac{1}{2.3}\);...;\(\frac{1}{2013^2}\)<\(\frac{1}{2012.2013}\)
\(\Rightarrow\)\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+...+\(\frac{1}{2013^2}\)<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+...+\(\frac{1}{2012.2013}\)=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2012}\)-\(\frac{1}{2013}\)=1-\(\frac{1}{2013}\)<1
Vậy bài toán đc chứng minh