For each positive integer $n$, let $x_n=1/(n+1)+1/(n+2)+\cdots+1/(2n)$. Prove that the sequence $(x_n)$ converges.

We write $$x_n=\sum_{k=1}^n\frac{1}{n+k}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+k/n}\to\int_0^1\frac{dx}{1+x}=\log 2$$

Hints:

$$\frac12=\frac n{2n}\le\frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}\le\frac n{n+1}$$

$$X_{n+1}:=\frac1{n+2}+\frac1{n+3}+\ldots+\frac1{2(n+1)}\le \frac1{n+1}+\frac1{n+2}+\ldots+\frac1{2n}=:X_n$$

The sequence is decreasing because $$x_{n+1}-x_n = \frac1{2n+2}+\frac1{2n+1} - \frac1{n+1} = \frac{1}{2(2n+1)(n+1)}\ge 0$$ and obviously, $x_n\le \frac{n}{n+1}\le 1$, so is convergent. For proving that $x_n\to \log 2$ the better idea is bounding $x_n$ by definite integrals of the form $\int_a^b\frac1x\,dx$ (how?).