a: Ta có: \(\frac{1}{101}<\frac{1}{100}\)
\(\frac{1}{102}<\frac{1}{100}\)
...
\(\frac{1}{200}<\frac{1}{100}\)
Do đó: \(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}<\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}\)
=>\(\frac{1}{101}+\frac{1}{102}+\cdots+\frac{1}{200}<\frac{100}{100}=1\) (ĐPCM)





