Giải thích các bước giải:
$A=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{36}+\dfrac{1}{64}+\dfrac{1}{100}+\dfrac{1}{144}+\dfrac{1}{196}+\dfrac{1}{256}+\dfrac{1}{324}\\
=\dfrac{1}{4}\left ( 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\right )\\
<\dfrac{1}{4}\left ( 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9} \right )\\
=\dfrac{1}{4}\left ( 1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{6}+\dfrac{1}{7}-\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9} \right )\\
=\dfrac{1}{4}\left ( 2-\dfrac{1}{9} \right )\\
=\dfrac{1}{4}\left ( \dfrac{18}{9}-\dfrac{1}{9} \right )\\
=\dfrac{1}{4}.\dfrac{17}{9}\\
=\dfrac{17}{36}<\dfrac{18}{36}=\dfrac{1}{2}\\
\Rightarrow A<\dfrac{1}{2}$
