Partial sums over rational functions

Question

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.

See What are the techniques to find the sum to infinity of a series?. There is not always a closed form, see Apéry's constant. — Vepir, Aug 22 '20 at 14:30
In general you cannot expect the partial sums of a rational function to have a closed form. e.g., even $\sum_{n=1}^k 1/n$ is the $k$-th harmonic number which does not to my knowledge have a closed form. So, in some sense this series is special, although you can generate examples like this by taking the differences $f(n+1) - f(n)$ for a given rational function $f$. — Jair Taylor, Aug 22 '20 at 15:10
As for techniques for finding the sum, maybe use partial fractions and hope it telescopes? — Jair Taylor, Aug 22 '20 at 15:11

score 4 · Accepted Answer · edited Aug 22 '20 at 17:40

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.

Partial sums over rational functions

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