\(A=\frac54+\frac{10}{9}+\cdots+\frac{2501}{2500}\)
\(=1+\frac14+1+\frac19+\ldots+1+\frac{1}{2500}\)
\(=49+\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}\right)\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{50^2}<\frac{1}{49\cdot50}=\frac{1}{49}-\frac{1}{50}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{49}-\frac{1}{50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1\)
=>\(0<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}<1\)
=>\(0+49<\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{50^2}+49<1+49\)
=>49<A<50
=>A không là số nguyên
