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?

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. 
