Cho (A = frac{1}{{{2^2}}} + frac{1}{{{3^2}}} + frac{1}{{{4^2}}} +  cdots  + frac{1}{{{{2013}^2}}}) .

* Hướng dẫn giải

Ta có: 

\(\frac{1}{{{3^2}}} < \frac{1}{{2.3}};\frac{1}{{{4^2}}} < \frac{1}{{3.4}};...\frac{1}{{{{2013}^2}}} < \frac{1}{{2012.2013}}\)

\(\begin{array}{l}
A < \frac{1}{{{2^2}}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{2012.2013}}\\
A < \frac{1}{4} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{2012.2013}}
\end{array}\)

Đặt \(B = \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{2012.2013}} = \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{{2012}} - \frac{1}{{2013}} = \frac{1}{2} - \frac{1}{{2013}}\)

\(\begin{array}{l}
 \Rightarrow A < \frac{1}{4} + \frac{1}{2} - \frac{1}{{2013}} = \frac{{6035}}{{8052}} < \frac{{6039}}{{8052}}\\
A < \frac{{6039}}{{8052}} = \frac{3}{4}
\end{array}\)

Vậy \(A < \frac{3}{4}\)