Ta có:
\(S=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)\(=\frac{1}{4}-\frac{1}{100}=\frac{24}{100}< \frac{50}{100}=\frac{1}{2}\)
Ta có : \(\frac{1}{5^2}=\frac{1}{5.5}< \frac{1}{4.5}\)
\(\frac{1}{6^2}=\frac{1}{6.6}< \frac{1}{5.6}\)
\(\frac{1}{7^2}=\frac{1}{7.7}< \frac{1}{6.7}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow S< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(S< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\)
\(S< \frac{1}{4}-\frac{1}{100}=\frac{6}{25}=\frac{24}{100}\)
Mà \(\frac{24}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow S< \frac{1}{2}\)
Vậy S<\(\frac{1}{2}\).
