A = $\frac{1}{2^{2}}$ + $\frac{1}{3^{2}}$ + ... + $\frac{1}{9^{2}}$
Có:
$\frac{1}{2^{2}}$ > $\frac{1}{2.3}$
$\frac{1}{3^{2}}$ > $\frac{1}{3.4}$
...
$\frac{1}{9^{2}}$ > $\frac{1}{9.10}$
A > $\frac{1}{2.3}$.$\frac{1}{3.4}$ ...$\frac{1}{9.10}$
= $\frac{1}{2}$ - $\frac{1}{10}$
= $\frac{2}{5}$
⇒ A > $\frac{2}{5}$ (1)
Lại có:
$\frac{1}{2^{2}}$ < $\frac{1}{1.2}$
$\frac{1}{3^{2}}$ < $\frac{1}{2.3}$
...
$\frac{1}{9^{2}}$ < $\frac{1}{8.9}$
A < $\frac{1}{1.2}$.$\frac{1}{2.3}$ ...$\frac{1}{8.9}$
= 1- $\frac{1}{9}$
= $\frac{8}{9}$
⇒ A < $\frac{8}{9}$ (2)
Từ 1 và 2 ⇒ $\frac{8}{9}$ > A > $\frac{2}{5}$
