b.Ta có:$U_{n+1}-U_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$ $=\frac{1}{2n+1}-\frac{1}{2n+2}$ $=\frac{1}{(2n+1)(2n+2)}>0$suy ra $U_n$ là dãy tăng thực sự.
b.Ta có:$U_{n
} = \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}$$U_{n+1} = \frac{1}{n+2} + \frac{1}{n+3} + ... + \frac{1}{2n+1}+\frac{1}{2n+2}$$U_{n+1}-U_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}$ $=\frac{1}{2n+1}-\frac{1}{2n+2}$ $=\frac{1}{(2n+1)(2n+2)}>0$suy ra $U_n$ là dãy tăng thực sự.