\(S=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}
\(1\) = \(1\)
\(\frac{1}{2^2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3^2}\) < \(\frac{1}{2.3}\)
.........
\(\frac{1}{100^2}\) < \(\frac{1}{99.100}\)
\(\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\) < \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
Ta có: \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1+1-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(\Rightarrow2-\frac{1}{100}\le2\)
Nên \(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\le2\)
=>\(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\le2\)
Vậy S \(\le2\)
