Limit of a sum that seems telescopable Calculate the limit:
$$\lim\limits_{n\to\infty}\sum\limits_{k=6}^{n}\frac{k^3-12k^2+47k-60}{k^5-5k^3+4k}$$ 
All I managed to do was to write the inside term as $\frac{(k-3)(k-4)(k-5)}{k(k+1)(k+2)(2-k)(1-k)}$ which leads me to think of telescoping but i dont know how. 
I think that comparing it to the analog product somehow makes sense or using the fact that $\frac{1}{(2-k)(1-k)} =  \frac{1}{1-k} - \frac{1}{2-k}$ helps.

Hint: Try to find the partial fraction decomposition of the rational function "inside". You will indeed find a sort of telescopic sum with a few first terms and then a few terms that tend to 0 when n tends to infinity 
Hint 2: $\frac{k^3−12k^2+47k−60}{k(k-1)(k+1(k-2)(k+2)}=\frac{\left(-15\right)}{k}+\frac{\frac{13}{3}}{k+1}+\frac{\frac{59}{3}}{k-1}+\frac{\left(-\frac{97}{12}\right)}{k-2}+\frac{\left(-\frac{11}{12}\right)}{k+2}$
then:  $\sum_{k=6}^n \frac{k^3−12k^2+47k−60}{k(k-1)(k+1(k-2)(k+2)} = \sum_{k=6}^n \frac{\left(-15\right)}{k}+\frac{\frac{13}{3}}{k+1}+\frac{\frac{59}{3}}{k-1}+\frac{\left(-\frac{97}{12}\right)}{k-2}+\frac{\left(-\frac{11}{12}\right)}{k+2}= 
(-15)\sum_{k=6}^n \frac{1}{k} + (13/3) \sum_{k=6}^n \frac{1}{k+1} + (59/3)\sum_{k=6}^n \frac{1}{k-1} 
+ (-97/12) \sum_{k=6}^n \frac{1}{k-2} + (-11/12) \sum_{k=6}^n \frac{1}{k+2}$
and I let you finish the job.
