b.Ta có:Un+1−Un=12n+1+12n+2−1n+1 =12n+1−12n+2 =1(2n+1)(2n+2)>0suy ra Un là dãy tăng thực sự.
b.Ta có:$U_{n
} = \frac{1}{n+1} + \frac{1}{n+2} + ... + \frac{1}{2n}Un+1=1n+2+1n+3+...+12n+1+12n+2U_{n+1}-U_n=\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n+1}
=\frac{1}{2n+1}-\frac{1}{2n+2}
=\frac{1}{(2n+1)(2n+2)}>0
suyraU_n$ là dãy tăng thực sự.