Question

[Thư Nguyễn Nguyễn]

Cho A = \(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{4026},B=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4025}\)

So sánh \(\dfrac{A}{B}\) với \(1\dfrac{2013}{2014}\)

Answer 1

[Phương Trâm]

Giải:

Ta có:

\(\dfrac{A}{B}=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{4026}}{1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{4025}}\)

\(\Rightarrow\dfrac{A}{B}=\dfrac{\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2046}\right)}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)

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

\(\Rightarrow\dfrac{A}{B}=1+\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2046}}{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{4025}}\)

Dễ thấy \(\dfrac{A}{B}>1\)

Mà \(\dfrac{2013}{2014}< 1\)

\(\Rightarrow\dfrac{A}{B}>1\dfrac{2013}{2014}\)