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