<s>[INST] Convergence of $\sum_{n=1}^{\infty}\left(\, \frac{1}{n} - \frac{1}{n + 2}\,\right)$ What criteria can I use to prove the convergence of
$$
\sum_{n=1}^{\infty}\left(\,{1 \over n} - {1 \over n + 2}\,\right)\ {\large ?}
$$
My idea was to use ratio test:
$$\displaystyle{1 \over n} - {1 \over n+2} = {2 \over n^{2} + 2n}$$
$$\displaystyle\frac{2}{\left(\, n + 1\,\right)^{2} + 2\left(\, n + 1\,\right)} \frac{n^{2} + 2n}{2} = \frac{n^{2} + 2n}{n^{2} + 4n + 3}$$
Of course $\displaystyle n^{2} + 2n \lt n^{2} + 4n + 3$ for all
$\displaystyle n$ , but
$$\displaystyle\lim \limits_{n \to \infty} \frac{n^{2} + 2n}{n^{2} + 4n + 3}=1$$
so I am not quite sure if I can apply ratio test.
 [/INST] Here's another way to prove the convergence 
$$ \sum \limits_{n=1}^{\infty} \left(\frac1n - \frac{1}{n+2}\right)= \sum \limits_{n=1}^{\infty} \frac{n+2-n}{n(n+2)} $$
$$= \sum \limits_{n=1}^{\infty} \frac{2}{n^2+2n}=2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n} $$
Also note that
$$ \left|\frac{1}{n^2+2n}\right|\leq \left|\frac{1}{n^2}\right| $$
And by the p-series test, we have
$$ \sum \limits_{n=1}^{\infty} \left|\frac{1}{n^2}\right| \Rightarrow \mbox{converges} $$
Which implies that
$$ \sum \limits_{n=1}^{\infty} \frac{1}{n^2} \Rightarrow \mbox{converges absolutely} $$
Therefore, by the direct comparison test 
$$ 2 \sum \limits_{n=1}^{\infty} \frac{1}{n^2+2n}  \Rightarrow \mbox{converges absolutely} $$
Absolute convergence implies convergence. Also a convergent series multiplied by $2$ is still a convergent series.
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