TA có: \(D=\frac34+\frac89+\cdots+\frac{1599}{1600}\)
\(=1-\frac14+1-\frac19+\cdots+1-\frac{1}{1600}\)
\(=1-\frac{1}{2^2}+1-\frac{1}{3^2}+\cdots+1-\frac{1}{40^2}\)
\(=39-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{40^2}\right)\)
Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)
\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)
...
\(\frac{1}{40^2}<\frac{1}{39\cdot40}=\frac{1}{39}-\frac{1}{40}\)
Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{40^2}<1-\frac12+\frac12-\frac13+\cdots+\frac{1}{39}-\frac{1}{40}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{40^2}<1-\frac{1}{40}<1\)
=>\(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{40^2}\right)>-1\)
=>\(-\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{40^2}\right)+39>-1+39\)
=>D>38
