<s>[INST] Summation of a rational function Calculate $\sum_{r=2}^{n} \frac{3r^2-1}{(r^3-r)^2}$.
My approach till now: $T_r=\frac{3r^2 -1}{r^2(r+1)^2(r-1)^2}$. Now let,  $$3r^2-1= a(r^2)+ b(r+1)^2 + c(r-1)^2$$
we find, $a=4$, $b=c=-\frac{1}{2}$. now we break it up, $$T_r = \frac{4}{(r+1)^2(r-1)^2}-\frac{1}{2(r)^2(r+1)^2}-\frac{1}{2(r)^2(r-1)^2}$$
but now the problem is I can't convert any of them into a "telescopable" format
another insight I had was to break it up in another way,
$$T_r= \frac{r^2 + (r^2-1)+(r^2-1)+1}{r^2(r^2-1)(r^2-1)}$$
after a bit of simplifying,
$$T_r= \frac{1}{(r^2-1)(r^2-1)} + \frac{2}{r^2(r^2-1)}+\frac{1}{r^2(r^2-1)(r^2-1)}$$
now two terms are solvable but how to do the third term maybe if you simplify it even more we get,
$$T_r = \frac{1}{(r^2-1)(r^2-1)} + \frac{2}{r^2(r^2-1)}+\frac{r^2-(r^2-1)}{r^2(r^2-1)(r^2-1)}$$
$$T_r=\frac{1}{(r^2-1)(r^2-1)} + \frac{2}{r^2(r^2-1)}+ \frac{1}{(r^2-1)^2}-\frac{1}{r^2(r^2-1)}$$
now rewriting, $(r^2-1)^2 = (r+1)^2(r-1)^2$, we get
$$T_r= \frac{2}{(r+1)^2(r-1)^2}+\frac{1}{r^2(r^2-1)}$$
but now how to solve the first term...
 [/INST] You can do standard partial fraction decomposition. That is, let $$\frac{3r^2-1}{(r^3-r)^2} = \frac{a}{r}+\frac{b}{r^2}+\frac{c}{r-1}+\frac{d}{(r-1)^2}+\frac{e}{r+1}+\frac{f}{(r+1)^2}.$$ Then you will get $$-\frac{1}{r^2}+\frac{1}{2(r-1)^2}+\frac{1}{2(r+1)^2}.$$ Can you do the telescoping from here and conclude?
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