We know that the series $\sum_{n=1}^\infty \dfrac{1}{n(n+1)}$ coverge to 1.
I want use Cauchy criterion to show the series converges as an exercise. Is the following proof correct?
Given $\epsilon >0$, we want to show that we can find an $N$ such that \begin{align*} \left| \sum_{n+1}^m \frac{1}{k(k+1)}\right|=\left|\frac{1}{n+1}-\frac{1}{m+1}\right| \le \epsilon \end{align*} whenever $m> n > N$.
Let $N=\dfrac{1}{\epsilon}$ then if $m > n \ge N$, we have \begin{align*} \left|\frac{1}{n+1}-\frac{1}{m+1}\right| \le \dfrac{1}{n+1} \le \dfrac{1}{N+1} \le \epsilon. \end{align*} Hence by the Cauchy criterion, the series is convergent.