Proof that $\sum_{n = 0}^{\infty}\dfrac{1}{n(n+1)}$ converges to $1$ I'm looking for some verification of my proof. The problem is asking me to prove that the sequence $\{s_{n}\}$ defined by 
$$s_{n} = \dfrac{1}{n(n + 1)}$$
converges to $1$.I thought it goes to $0$, but maybe I'm missing something. Here's the problem word-for-word:
Problem
Define the sequence $\{s_{n}\}$ by 
$$s_{n} = \dfrac{1}{2 \cdot 1} + \dfrac{1}{3 \cdot 2} + \ldots + \dfrac{1}{(n + 1)(n)}$$ for every index $n$.
Prove that the limit as $n$ approaches infinity of $s_{n}$ equals $1$.
My proof:
Let $\epsilon > 0$ be arbitrarily chosen. Choose $M = \dfrac{1}{\epsilon}$, and let $n \geq M$ be arbitrarily chosen. Then,
$$\left| \dfrac{1}{n(n + 1)} - 1 \right| < \left| \dfrac{1}{n(n + 1)}\right| < \left|\dfrac{1}{n}\right| = \dfrac{1}{n} < \epsilon.$$
(I was able to remove the absolute value because $n > 0 ,$ always.)
Equivalently, this means that 
$$n > \dfrac{1}{\epsilon}.$$
Thus, choosing $M = \dfrac{1}{e}$ means that the above inequality always holds, which completes the proof.
Updated Proof
Ok, so I think I was actually supposed to prove 
$$\sum_{n = 0}^{\infty} \dfrac{1}{n(n + 1)} = 1.$$
By applying partial fraction decomposition, I found that this summation equals
$$\sum_{n = 0}^{\infty}\dfrac{1}{n} - \sum_{n = 0}^{\infty}\dfrac{1}{n + 1}.$$
From here, I can clearly see that everything cancels out except for the $1$ term. But is it possible to show this using the definition of convergence?

We have that
$$0\le s_{n} = \dfrac{1}{n(n + 1)}\le \frac1n\to 0$$
and
$$\sum_1^\infty s_{n} = \sum_1^\infty \dfrac{1}{n(n + 1)}=\sum_1^\infty \left(\dfrac{1}{n}-\dfrac{1}{n + 1}\right) =1\color{red}{-\frac12+\frac12-\frac13+\frac13-\frac14+\frac14-}\ldots$$
note also that
$$S_k=\sum_1^k s_{n}=1-\frac1{k+1} \to 1$$
