Mạn phép sửa lại đề : \(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\)
Ta nhận thấy : \(\dfrac{1}{k^2}=\dfrac{1}{k.k}< \dfrac{1}{k.\left(k-1\right)}\) Vì : k > k - 1
Lại có : \(\dfrac{1}{k\left(k-1\right)}=\dfrac{1}{k-1}-\dfrac{1}{k}\)
Ta có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\) < \(\dfrac{1}{2\left(2-1\right)}+\dfrac{1}{3\left(3-1\right)}+...+\dfrac{1}{n\left(n-1\right)}\)
\(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\) < \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
Do : n > 1 , nên : \(\dfrac{1}{2^2}+\dfrac{1}{3^3}+...+\dfrac{1}{n^2}\) < 1
@Phùng Khánh Linh Đề này đúng rồi mà bạn, không cần sửa đâu
Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{n^2}\)
\(\dfrac{1}{2}A=\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2n^2}\)
\(A-\dfrac{1}{2}A=\dfrac{1}{2^2}-\dfrac{1}{2n^2}\)
\(\dfrac{1}{2}A=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n^2}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{n^2}\)
Vì \(\dfrac{1}{2}-\dfrac{1}{n^2}< \dfrac{1}{2}\)
Mà \(\dfrac{1}{2}< 1\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{n^2}< 1\)
\(\Rightarrow\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{n^2}< 1\left(Đpcm\right)\)
