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Real Madrid

14/04/2017 at 21:33
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Prove that :

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

 




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    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)


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