Question

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Answer 1

\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(\dfrac{1}{1}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)

=>\(A< \dfrac{1}{2^2}+\dfrac{1}{2^2}\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=\dfrac{1}{2^2}+\dfrac{1}{2^2}\cdot\dfrac{49}{50}=\dfrac{99}{200}< \dfrac{1}{2}\)