I recently came across the result that
$$\sum_{n=2}^\infty \frac{n^4-n^3+n+1}{n^6-1} = \frac{1}{2}$$
I am wondering how one could proof this, generally how one could evaluate a sum over rational functions.
If I plug the sum into Wolfram Alpha it gives
$$\frac{3k^4-k-2}{6k(k+1)(k^2+k+1)}$$
as the $k$-th partial sum. Taking the limit as $n \to \infty$, this would in fact proof the upper equality.
Sadly, I could not wrap my head around how to get to Wolfram Alphas partial sum result. If anyone has an idea let me know. Any tips are appreciated.
Best Answer
In a word: the terms telescope in a nice way.
Using partial fractions, we have $$ \frac{n^4-n^3+n+1}{n^6-1} = \frac{1}{3}\left(\frac{n - 2}{n^2 - n + 1}- \frac{n-1}{n^2 + n + 1} + \frac{1}{n - 1} - \frac{1}{n + 1}\right) $$We need to be a bit careful because the harmonic series diverges, so we should only group terms with opposite signs together. The last two terms form a well-known telescoping series: $$ \frac{1}{3}\sum_{n=2}^{m}\left(\frac{1}{n - 1} - \frac{1}{n + 1} \right)= \frac{1}{3}\left(1+\frac{1}{2}-\frac{1}{m}-\frac{1}{m+1}\right)\longrightarrow \frac{1}{2} $$The first two terms telescope as well, with the second term eating the one before it: $$ \frac{1}{3}\sum_{n=2}^{m}\left(\frac{n - 2}{n^2 - n + 1}- \frac{n-1}{n^2 + n + 1} \right)= -\frac{1}{3}\cdot \frac{m-1}{m^2+m+1}\longrightarrow 0 $$This sort of thing won't work in general but I won't look a gift horse in the mouth.