Ta có:P=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)=\(\frac{1}{2.2}+\frac{1}{3.3}+...+\frac{1}{n.n}\)
<\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right).n}\)=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
=\(\frac{1}{1}-\frac{1}{n}=\frac{n}{n}-\frac{1}{n}=\frac{n-1}{n}\)<1
=>P<1