Claim:The series $\displaystyle \sum_{n=1}^{ \infty} \frac{2}{n^2 + n}$ converges.

By definition:

The series $\displaystyle \sum_{n=1}^{ \infty} \frac{2}{n^2 + n} = \sum_{n=1}^{ \infty} a_n$ converges if and only if for every $\displaystyle \epsilon > 0$, there exists some $\displaystyle N \in \mathbb{N}$ such that for all $\displaystyle n, m$ with $\displaystyle n > m \geq N$, we have $\displaystyle |a_{m+1} + a_{m+2}+ ...+ a_n| < \epsilon$.

(Attempted) proof:

Let $\displaystyle m \in \mathbb{N}$ be arbitrary but fixed, and let $\displaystyle S_m = \sum_{n=1}^{m} \frac{2}{n^2 + n}$. Then the sequence $\displaystyle (S_m)$ is convergent, and it necessarily follows that $\displaystyle (S_m)$ is Cauchy. That is, for all $\displaystyle \epsilon > 0$, there exists some $\displaystyle N \in \mathbb{N}$ such that for all $\displaystyle n, m \geq N$, we have $\displaystyle |S_{n} - S_{m}| < \epsilon$.

But if $\displaystyle n > m \geq N$, we have $\displaystyle |S_n - S_m| = |a_{m+1} + a_{m+2}+ ...+ a_n| < \epsilon$

which shows, directly from the definition, that the series $\displaystyle \sum_{n=1}^{ \infty} \frac{2}{n^2 + n}$ is convergent. $\displaystyle \square$

Is this satisfactory? Or is there a more simplistic approach that I should be taking?