Ta có:
$\frac{1}{{{3}^{2}}}=\frac{1}{3.3}>\frac{1}{3.4}$
$\frac{1}{{{4}^{2}}}=\frac{1}{4.4}>\frac{1}{4.5}$
…
$\frac{1}{{{100}^{2}}}=\frac{1}{100.100}>\frac{1}{100.101}$
Khi đó:
$\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+...+\frac{1}{{{100}^{2}}}>\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}$
$\Rightarrow \frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+...+\frac{1}{{{100}^{2}}}>\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{100}-\frac{1}{101}$
$\Rightarrow \frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+...+\frac{1}{{{100}^{2}}}>\frac{1}{3}-\frac{1}{101}$
$\Rightarrow \frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+...+\frac{1}{{{100}^{2}}}>\frac{1}{4}+\frac{1}{3}-\frac{1}{101}$
$\Rightarrow \frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+...+\frac{1}{{{100}^{2}}}>\frac{695}{1212}>\frac{1}{2}$