How to evaluate $\sum\limits_{k=0}^{n}\arctan f(k)$ where $f(k)$ is a rational fraction Find the sum  closed form 
$$\sum_{k=0}^{n}\arctan{\dfrac{k^4+6k^3+10k^2-k-9}{(k+1)(k+2)(k+3)(k^3+7k^2+15k+8)}}$$
For problems involving sums, the idea is to use trigonometricidentities to write the  sum in the form
$$\sum_{k=1}^{n}[g(k)-g(k-1)]$$
and I initially considered pairing every two terms up to use the $arctanx+arctany $ trick, but it doesn't work because each arctanarctan term has a different coefficient.

By setting $f(k)=3+(7+2i)k+(5+i)k^2+k^3$ we are dealing with:
$$ \sum_{k=1}^{n}\text{arg}\left(f(k)\;\overline{f(k+1)}\right) =\text{arg }f(1)-\text{arg }f(n+1).\tag{1}$$
In order to notice that, I factored $(k+1)(k+2)(k+3)(k^3+7k^2+15k+8)+i(k^4+6k^3+10k^2-k-9)$, getting:
$$\left(3+(7+2 i) k+(5+i) k^2+k^3\right) \left((16-3 i)+(20-4 i) k+(8-i) k^2+k^3\right)\tag{2}$$
and checked that $(2)$ has the $f(k)\,\overline{f(k+1)}$ structure.
