Question

Answer 1

b: \(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)

=>\(S< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{8\cdot9}\)

=>\(S< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}=1-\dfrac{1}{9}=\dfrac{8}{9}\)

\(S=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)

=>\(S>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)

=>\(S>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

=>\(S>\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{5-1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

=>\(\dfrac{2}{5}< S< \dfrac{8}{9}\)

a: