Ta có: \(\frac{1}{2^2}\)< \(\frac{1}{1.2}\)
\(\frac{1}{3^2}\)< \(\frac{1}{2.3}\)
\(\frac{1}{4^2}\) < \(\frac{1}{3.4}\)
....
\(\frac{1}{100^2}\)<\(\frac{1}{99.100}\)
=> \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\) < \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
Ta có: \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(1-\frac{1}{100}\)
Vì \(1-\frac{1}{100}\) < 1
Nên\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{99.100}\) < 1
Vậy \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{100^2}\) < 1
