Question

[Hoàng Thu Huyền]

Chứng minh rằng:

a, M=1/2^2 + 1/3^2 + 1/4^2 +..+ 1/n^2 <1

Answer 1

[Komorebi]

Ta có : \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)

\(\Rightarrow M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)

\(\Rightarrow M< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(\Rightarrow M< 1-\dfrac{1}{n}< 1\)

Vậy \(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\)

Answer 2

[Đỗ Vũ Bá Linh]

Để \(M< 1\), ta phải có điều kiện: \(n\in\) R*. Nếu \(n=0\) thì \(M\) không xác định.
\(M=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)
                                                 \(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
                                                 \(=1-\dfrac{1}{n}< 1\)
Vậy \(M< 1\) với \(n\in\) R*.