Sequence and Series bounded Problem: Let $ x_n $ be a sequence defined by $x_n = \displaystyle {\sum_{k=n+1}^{ 2n}\frac{1}{k}}$. Show that $ x_n$ converge. 
I can bounded $ x_n <\frac{ n}{ n+1}$, but I can't show that $ x_n$ is increasing, thus would use that es increasing and bounded then is convergent.

Simply notice that, since $\frac{1}{2n+1} > \frac{1}{2n+2}$, $$x_{n+1} - x_n = \sum_{k=n+2}^{2n+2} \frac{1}{k} - \sum_{k=n+1}^{2n} \frac{1}{k} = \frac{1}{2n+1}  + \frac{1}{2n+2} - \frac{1}{n+1} > \frac{2}{2n+2} - \frac{1}{n+1} = 0.$$
Hence the sequence is increasing. Then apply your bound $x_n \le \frac{n}{n+1} < 1$ to conclude convergence. 
