Question

Answer 1

Ta có: \(A=\frac12+\frac{1}{2^2}+\cdots+\frac{1}{2^{2021}}\)

=>\(2A=1+\frac12+\cdots+\frac{1}{2^{2020}}\)

=>2A-A=\(1+\frac12+\cdots+\frac{1}{2^{2020}}-\frac12-\frac{1}{2^2}-\cdots-\frac{1}{2^{2021}}\)

=>\(A=1-\frac{1}{2^{2021}}\)

=>A<1

Ta có: \(B=\frac13+\frac14+\frac15+\frac{13}{60}\)

\(=\frac{20}{60}+\frac{15}{60}+\frac{12}{60}+\frac{13}{60}\)

\(=\frac{60}{60}\)

=1

=>A<B