Question

[Trần Thị Trà My]

Cho A=\(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Chứng minh A<2

Answer 1

[Hoang Hung Quan]

Ta có:

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)

\(=1+\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\right)\)

Đặt \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\) ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)

\(.....................\)

\(\dfrac{1}{50^2}=\dfrac{1}{50.50}< \dfrac{1}{49.50}\)

Cộng các vế trên với nhau ta được:

\(B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)

\(\Rightarrow B< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(\Rightarrow B< 1-\dfrac{1}{50}< 1\)

\(\Rightarrow1+B< 1+1=2\) Hay \(A< 2\)

Vậy \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 2\) (Đpcm)

Answer 2

[Mới vô]

\(A=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ =1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\\ \Rightarrow A< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}=1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=1+1-\dfrac{1}{50}=2-\dfrac{1}{50}< 2\)