What is $\sum_{n=1}^k\frac{1}{n^3-c}$? I am currently doing a proof which requires me, given some constant $c$ and some constant $k$, to evaluate the following:
$$\sum_{n=1}^k\frac{1}{n^3-c}$$
Is there any known way to do such?
If so, can it be generalized to:
$$\sum_{n=1}^k\frac{1}{n^m-c}$$
Wolfram Alpha gives me a close form expression (for the former equation), but doesn't explain itself. It would be nice of one could give me an explanation of their reasoning.
 A: $$S(c,m,k)=\sum_{n=1}^k\frac{1}{n^m-c}\qquad (m\text{ integer})$$
FIRST : Solve the equation $\quad\nu^m=c\quad$ for $\nu$.
They are $m$ roots (real real and complex), namely $\quad\nu_j\quad$ with $\quad 1\leq j \leq m$.
SECOND : Convert $\quad\frac{1}{n^m-c}\quad$ to partial fractions : 
$$\frac{1}{n^m-c}=\sum_{j=1}^m \frac{a_j}{n-\nu_j}$$
Compute explicitly the $m$ coefficients $a_j$.
THIRD : Express $\quad\sum_{n=1}^k\frac{1}{n-\nu_j}\quad$ on closed form :
In the general case (arbitrary $\nu_j$) it is not possible with a finite number of elementary function. This requires a special function, namely the digamma function $\psi(x)$ :
http://mathworld.wolfram.com/DigammaFunction.html
A property of this function is : $\quad\sum_{n=1}^k\frac{1}{n-C}=\psi(m+1-C)-\psi(1-C)\quad$. 
Thus, for each $\nu_j$ :
$$\sum_{n=1}^k\frac{1}{n-\nu_j}=\psi(m+1-\nu_j)-\psi(1-\nu_j)$$
RESULT :
$$\sum_{n=1}^k\frac{1}{n^m-c} = \sum_{n=1}^k a_j\left(\psi(m+1-\nu_j)-\psi(1-\nu_j) \right)$$
You can do it in the case $\sum_{n=1}^k\frac{1}{n^3-c}$ : Compute $\nu_1\:,\:\nu_2\:,\:\nu_3$ , then $a_1\:,\:a_2\:,\:a_3$.
NOTE : For some particular values of $m$ the function digamma can be reduced to elementary functions (for example $m=4$). As a matter of fact, in those particular cases, the conversion to partial fractions leads then directly to the solution with no need for the function digamma.
