Question

Cho A=1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4026}\)

        B=1+\(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4025}\)

So sánh \(\frac{A}{B}vs1\frac{2013}{2014}\)

Answer 1

Ta có

\(\frac{A}{B}=\frac{1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)

\(\Rightarrow\frac{A}{B}=\frac{\left(1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}\right)+\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{4026}\right)}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)

\(\Rightarrow\frac{A}{B}=\frac{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}+\frac{\frac{1}{2}+\frac{1}{4}+....+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)

\(\Rightarrow\frac{A}{B}=1+\frac{\frac{1}{2}+\frac{1}{4}+....+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+....+\frac{1}{4025}}\)

Dễ thấy A/B > 1

2013/2014<1

=> \(\frac{A}{B}>\frac{2013}{2014}\)