How do I solve $\sum_{n=5}^\infty \frac{12}{16 \, n^{2} + 40 \, n + 21}$ How do I solve 
$\displaystyle\sum_{n=5}^\infty \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}$ ?
A tip was to turn this into a telescoping series but I don't see how I could do this. Are there any guidelines in how I could transform this series into a telescoping series?
 A: Proceed like this. 
First observe the denominator which is $16n^2+40n+21$. If you factorise , you will get as $(4n+3)(4n+7)$. 
Next, write $\frac{12}{16n^2+40n+21}$ as $\frac{A}{4n+3}+\frac{B}{4n+7}$ where $A$ and $B$ are to be determined. If you know how to do so then it is fine else let me know. 
The values of $A, B$ are $3, -3$ respectively. 
So we shall get $\frac{12}{16n^2+40n+21}=\frac{3}{4n+3}-\frac{3}{4n+7}=\frac{3}{4n+3}-\frac{3}{4(n+1)+3}$.
Rest is easy.
Most of time partial fraction decomposition works like this. 
A: $\begin{align} \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}&=\displaystyle\frac{12}{(4n+3)(4n+7)}\\&=3\frac{4}{(4n+3)(4n+7)}\\&=\displaystyle3{\huge[}\frac{1}{(4n+3)}-\frac{1}{(4n+7)}{\huge]}\end{align}$
$\begin{align}\therefore \displaystyle\sum_{n=5}^\infty \displaystyle\frac{12}{16 \, n^{2} + 40 \, n + 21}&=3\displaystyle\sum_{n=5}^\infty \displaystyle\frac{1}{(4n+3)}-\frac{1}{(4n+7)}\\&=3{\huge[}\frac{1}{23}-\frac{1}{27}+\frac{1}{27}-\frac{1}{31}+\cdots+\\&\frac{1}{(4n+3)}-\frac{1}{(4n+7)}+\frac{1}{(4n+7)}-\frac{1}{(4n+11)}+\cdots{\huge]}\\&=\frac{3}{23}\end{align}$
