Giải thích các bước giải:
a.Ta có:
$A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}$
$\to A=\dfrac{1}{2.2}+\dfrac1{3.3}+\dfrac{1}{4.4}+...+\dfrac1{n.n}$
$\to A<\dfrac{1}{1.2}+\dfrac1{2.3}+\dfrac{1}{3.4}+...+\dfrac1{(n-1).n}$
$\to A<\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{n-(n-1)}{(n-1).n}$
$\to A<1-\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+....+\dfrac1n-\dfrac1{n-1}$
$\to A<1-\dfrac1{n-1}<1$
$\to A<1$
b.Ta có:
$B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{(2n)^2}$
$\to 4B=1+\dfrac1{2^2}+\dfrac1{3^2}+...+\dfrac{1}{n^2}$
Mà $\dfrac1{2^2}+\dfrac1{3^2}+...+\dfrac{1}{n^2}<1$(câu a)
$\to 4B<1+1$
$\to 4B<2$
$\to B<\dfrac12$