Compute $ \lim_{n\rightarrow \infty }\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k} $ Calculate $ \lim_{n\rightarrow \infty }\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k} $.
So far I found that $ \frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=\frac{(k-5)(k-4)(k-3)}{(k-2)(k-1)k(k+1)(k+2)}=\frac{((k-3)!)^{2}}{(k-6)!\cdot (k+2)!}. $
 A: You did a big part of the job. 
$k^{5}-5k^{3}+4k=(k-2)(k-1)k(k+1)(k+2)$
Now you can rewrite $$\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=\frac{A}{k-1}+\frac{B}{k}+\frac{C}{k+1}+\frac{D}{k-2}+\frac{E}{k+2}$$.
By multiplying the numerators and denominators of the RHS, and identifying the new numerator in terms of $A,B,C...$ to the numerator of the lHS.
You will get 
$$\sum_{k=6}^{n}\frac{k^{3}-12k^{2}+47k-60}{k^{5}-5k^{3}+4k}=4\sum_{k=5}^{n-1}{\frac{1}{k}}-15\sum_{k=6}^{n}{\frac{1}{k}}+20\sum_{k=7}^{n+1}{\frac{1}{k}}-\frac{35}{4}\sum_{k=8}^{n+2}{\frac{1}{k}}-\frac{1}{4}\sum_{k=4}^{n-2}{\frac{1}{k}}$$
or equivalently
$$4\frac{1}{5}-20\frac{1}{6}+\frac{35}{4}(\frac{1}{6}+\frac{1}{7})-\frac{1}{4}(\frac{1}{4}+\frac{1}{5})+(4-15+20-\frac{35}{4}-\frac{1}{4})\sum_{k=6}^{n}{\frac{1}{k}}-4*\frac{1}{n}+20\frac{1}{n+1}-\frac{35}{4}(\frac{1}{n+2}+\frac{1}{n+1})+\frac{1}{4}(\frac{1}{n-1}+\frac{1}{n})$$
We have $$(4-15+20-\frac{35}{4}-\frac{1}{4})=0$$, therfore the limit is $$4\frac{1}{5}-20\frac{1}{7}+\frac{35}{4}(\frac{1}{8}+\frac{1}{7})-\frac{1}{4}=\frac{1}{16}$$
