Question

Cho \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\) . CMR A<1

Answer 1

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)\cdot n}=\dfrac{1}{n-1}-\dfrac{1}{n}\)

Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

=>\(A< 1-\dfrac{1}{n}\)

=>A<1

=>0<A<1

=>A không là số nguyên