<s>[INST] 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
 [/INST] 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 $\frac{7}{12}$.
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