Real Madrid
14/04/2017 at 21:33-
Ngu Ngu Ngu 15/04/2017 at 08:04
Put \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\)
We see:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)
\(\Rightarrow A< 1-\dfrac{1}{n}< 1\)
Conclude:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) (The thing must prove)