\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{20^2} \\>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{20.21} \\=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{20}-\dfrac{1}{21} \\=\dfrac{1}{2}-\dfrac{1}{21}=\dfrac{19}{42}>\dfrac{18}{42}=\dfrac{3}{7}\)