Question

[VTTH]

A=1/2^2+1/3^2+1/4^2+.......+1/63^2   Chứng minh rằng:A >1/2

Answer 1

[Nguyễn Lê Phước Thịnh]

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

\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}=\dfrac{1}{3}-\dfrac{1}{4}\)

...

\(\dfrac{1}{63^2}>\dfrac{1}{63\cdot64}=\dfrac{1}{63}-\dfrac{1}{64}\)

Do đó: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{63^2}>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{63}-\dfrac{1}{64}\)

=>\(A>\dfrac{1}{2}-\dfrac{1}{64}>\dfrac{1}{2}\)