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
 A: 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 $\lim_{n\to\infty}s_n=\frac{1}{3}+\frac{1}{4}$. 
A: Use partial fractions: Suppose
$$f(n)=\sum\limits_{k=3}^n\dfrac{2}{k^2+2k}=\sum\limits_{k=3}^n(\dfrac{1}{k}-\dfrac{1}{k+2})$$
$$=(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{5}-\frac{1}{7})+\dots+(\frac{1}{n-2}-\frac{1}{n})+(\frac{1}{n-1}-\frac{1}{n+1})+(\frac{1}{n}-\frac{1}{n+2})$$
All the terms cancel out except four:
$$f(n)=\displaystyle \frac{1}{3}+\frac{1}{4}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{7}{12}-\frac{2n+3}{(n+1)(n+2)}$$
Obviously the last fraction converges to zero and you 're left with $\frac{7}{12}$.
A: Hint: $ \dfrac{2}{k^2+2k}=\dfrac{(k+2)-k}{k(k+2)}=\dfrac{1}{k}-\dfrac{1}{k+2}$  
So, $\sum\limits_{k=3}^\infty \dfrac{2}{k^2+2k}=\bigg( \dfrac13-\dfrac15\bigg)+\bigg(\dfrac14-\dfrac16\bigg)+\bigg(\dfrac15-\dfrac17\bigg)+\ldots$  
Does this help?
