What converges this series? What exactly converges the series?
$\sum _{k=3}^{\infty \:}\frac{2}{k^2+2k}$
I tried taking out the constant
$=2\sum _{k=3}^{\infty \:}\frac{1}{k^2}$
then $p=2,\:\quad \:p>1\quad \Rightarrow \sum _{k=3}^{\infty \:}\frac{1}{k^2}$
but I really don't know what i'm doing

Note that $\dfrac{2}{k^2+2k}=\dfrac{1}{k}-\dfrac{1}{k+2}$. So our sum is 
$$\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+  \left(\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{8}-\frac{1}{10}\right)+\cdots.$$Remove the parentheses, and note the mass cancellations (telescoping). Almost everything dies, and we are left with $\dfrac{1}{3}+\dfrac{1}{4}$.
Remark: The above argument was informal. It can be replaced by a formal argument in which we take the sum $s_n$ of the first $n$ terms, and show that $\lim_{n\to\infty}s_n=\frac{1}{3}+\frac{1}{4}$. 
