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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.

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    Certainly $\le \epsilon$. You might want to say $\lt$, though of course $\le$ is perfectly correct. Good. Perhaps should make telescoping argument explicit. You know and I know why the collapse, but someone reading it might not.2012-12-06

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