Đáp án:
$A= \frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+...+\frac{1}{n^{2}}$
$\Rightarrow A< \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+...+\frac{1}{n\left ( n-1 \right )}$
$\Rightarrow A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}$
$\Rightarrow A< 1-\frac{1}{n}$
$\Rightarrow A< 1$
$B= \frac{1}{2^{2}}+\frac{1}{4^{2}}+\frac{1}{6^{2}}+...+\frac{1}{\left ( 2n \right )^{2}}$
$B< \frac{1}{4}+\frac{1}{2\cdot 4}+\frac{1}{4\cdot 6}+...+\frac{1}{\left ( 2n-2 \right )\cdot \left ( 2n \right )}$
$B< \left ( \frac{2}{4}+\frac{2}{2\cdot 4}+\frac{2}{4\cdot 6} +...+\frac{2}{\left ( 2n-2 \right )2n}\right )\cdot \frac{1}{2}$
$B< \left ( \frac{2}{4}+\frac{2}{4}-\frac{1}{2n} \right )\cdot \frac{1}{2}$
$B< \frac{1}{2}\Rightarrow B< 1$.
