\(B=\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{n^2-1}{n^2}\)
\(=1-\dfrac{1}{4}+1-\dfrac{1}{9}+...+1-\dfrac{1}{n^2}\)
\(=\left(1+1+...+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
\(=\left(n-2+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\)
\(\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{2\cdot3}< \dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)
\(\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{3\cdot4}< \dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)
...
\(\dfrac{1}{n-1}-\dfrac{1}{n}=\dfrac{1}{n\left(n-1\right)}< \dfrac{1}{n^2}< \dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Do đó: \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)
=>\(\dfrac{1}{2}-\dfrac{1}{n}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1-\dfrac{1}{n+1}\)
=>\(\dfrac{1}{2}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1\)
=>\(0< \dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< 1\)
=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\) không là số nguyên
=>\(-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\notin Z\)
=>\(B=\left(n-1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}\right)\notin Z\)
=>B không là số nguyên
