Chứng minh 2/5 < A < 8/9 với A=1/2^2+1/3^2+1/4^2+...+1/9^2

2. Chứng tỏ:\(\dfrac{2}{5}< A< \dfrac{8}{9}.\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)

Giải:

Ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)

\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}.\)

\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}.\)

\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}.\)

\(A< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}.\)

\(A< 1+0+0+0+...+0-\dfrac{1}{9}.\)

\(A< 1-\dfrac{1}{9}.\)

\(A< \dfrac{8}{9}_{\left(1\right)}.\)

Ta lại có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)

\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}.\)

\(A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}.\)

\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}.\)

\(A>\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-\dfrac{1}{10}.\)

\(A>\dfrac{1}{2}+0+0+0+...+\dfrac{1}{10}.\)

\(A>\dfrac{1}{2}-\dfrac{1}{10}.\)

\(A>\dfrac{4}{10}.\)

\(\Rightarrow A>\dfrac{2}{5}_{\left(2\right)}.\) (vì \(\dfrac{4}{10}=\dfrac{2}{5}.\))

Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\).

\(\Rightarrow A< \dfrac{8}{9}\) và \(A>\dfrac{2}{5}.\)

\(\Rightarrow\) \(\dfrac{8}{9}>A>\dfrac{2}{5}\) hay \(\dfrac{2}{5}< A< \dfrac{8}{9}.\)

Vậy ta thu được \(đpcm.\)

~ Học tốt!!!... ~ ^ _ ^