<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] How about computing the partial sums? For any $N>2$ we have:
$$\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right) = \frac{3}{2}-\frac{1}{N+1}-\frac{1}{N+2}$$
hence:
$$\left|\frac{3}{2}-\sum_{n=1}^{N}\left(\frac{1}{n}-\frac{1}{n+2}\right)\right|\leq\frac{2}{N}$$
ensures convergence (towards $\frac{3}{2}$).
</s>