Question

Prove 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dots+\dfrac{1}{n^2}<\dfrac{7}{4}

Steven Zheng Steven Zheng posted 1 week ago


Answer

1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dots+\dfrac{1}{n^2}

< 1+\dfrac{1}{2^2-1}+\dfrac{1}{3^2-1}+\dots+\dfrac{1}{n^2-1}

=1+\dfrac{1}{2}(\dfrac{1}{2-1}-\dfrac{1}{2+1}+\dfrac{1}{3-1}-\dfrac{1}{3+1}+\dots+\dfrac{1}{n-1}-\dfrac{1}{n+1})

1+\dfrac{1}{2}[(\dfrac{1}{2-1}+\dfrac{1}{3-1}) -(\dfrac{1}{n}+\dfrac{1}{n+1})]

< 1+\dfrac{1}{2}(\dfrac{1}{2-1}+\dfrac{1}{3-1})

= \dfrac{7}{4}

Steven Zheng Steven Zheng posted 1 week ago

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