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.
 A: I would write out the first few terms and see how they cancel.
A: 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 $\frac{3}{2}$).
A: You have $\frac{2}{n^2+n} \le \frac{2}{n^2}$. Thus you can use the direct comparision test to prove the convergence (if you already have proven in your course that $\sum_{n=1}^\infty \frac 1{n^2}$ and thus also $\sum_{n=1}^\infty \frac 2{n^2}$ converges).
Answer to your 2nd question: You never can apply the ratio test if the limit is 1 (as in this case).
A: 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.
A: You can use the limit comparison test. Let $a_{n} = \frac{2}{n^{2}+2n}$ and consider $b_{n} = \frac{1}{n^{2}}$ then 
$$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}= \lim_{n\rightarrow \infty}\frac{2n^{2}}{n^{2}+2n} = 2$$
hence $\sum a_{n}$ and $\sum b_{n}$ converge or diverge together.. but $b_{n}$ is a convergent $p$-series.
That being said, my up vote is for Mark Bennet.
A: How about the comparison test?
