Có:
[tex]\frac{1}{2^2} + \frac{1}{4^2} + ... + \frac{1}{(2n)^2}=\frac{1}{2^2}.(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2})<\frac{1}{2^2}.(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{(n-1)n})=\frac{1}{2^2}(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n})=\frac{1}{2^2}(2-\frac{1}{n})<\frac{1}{2^2}.2=\frac{1}{2}[/tex]