Question

Chứng tỏ:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{10}{11}\)

Answer 1

Ta có:

 A = \(\frac{1}{2^2}\) + \(\frac{1}{3^2}\) + \(\frac{1}{4^2}\)+....+ \(\frac{1}{11^2}\) 

A  = \(\frac{1}{2.2}\) + \(\frac{1}{3.3}\) + \(\frac{1}{4.4}\)+....+ \(\frac{1}{11.11}\)

A   < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +\(\frac{1}{3.4}\) + .... + \(\frac{1}{10.11}\)

A   < 1 - \(\frac{1}{2}\)+ \(\frac{1}{2}\) - \(\frac{1}{3}\) + \(\frac{1}{3}\) - \(\frac{1}{4}\) + ...... + \(\frac{1}{10}\) - \(\frac{1}{11}\)

A   <  1 - \(\frac{1}{11}\)

\(\Rightarrow\) A  <   \(\frac{10}{11}\)

Answer 2

Ta có : \(\frac{1}{2^2}< \frac{1}{1.2}\)

            \(\frac{1}{3^2}< \frac{1}{2.3}\)

             .........

              \(\frac{1}{11^2}< \frac{1}{10.11}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\)

Lại có : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)

\(=1-\frac{1}{11}\)

\(=\frac{10}{11}\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}=\frac{10}{11}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+..+\frac{1}{11^2}< \frac{10}{11}\) ( đpcm )