jmc

First, we decompose $\frac{4n^3-n^2-n+1}{n^6-n^5+n^4-n^3+n^2-n}$ into partial fractions. We factor the denominator: $$\begin{array}{rl}{n}^6-n^5+n^4-n^3+n^2-n& =n(n^5-n^4+n^3-n^2+n-1)\\ & =n(n^4(n-1)+n^2(n-1)+(n-1))\\ & =n(n-1)(n^4+n^2+1)\\ & =n(n-1)[(n^4+2n^2+1)-n^2]\\ & =n(n-1)[(n^2+1{)}^2-n^2]\\ & =n(n-1)(n^2+n+1)(n^2-n+1).\end{array}$$Then by partial fractions, $$\frac{4n^3-n^2-n+1}{n(n-1)(n^2+n+1)(n^2-n+1)}=\frac{A}{n}+\frac{B}{n-1}+\frac{Cn+D}{n^2+n+1}+\frac{En+F}{n^2-n+1}$$for some constants $A,$ $B,$ $C,$ $D,$ $E,$ and $F.$

Multiplying both sides by $n(n-1)(n^2+n+1)(n^2-n+1),$ we get $$\begin{array}{rl}4n^3-n^2-n+1& =A(n-1)(n^2+n+1)(n^2-n+1)\\ & +Bn(n^2+n+1)(n^2-n+1)\\ & +(Cn+D)n(n-1)(n^2-n+1)\\ & +(En+F)n(n-1)(n^2+n+1).\end{array}$$Setting $n=0,$ we get $-A=1,$ so $A=-1.$

Setting $n=1,$ we get $3B=3,$ so $B=1.$ The equation above then becomes $$\begin{array}{rl}4n^3-n^2-n+1& =-(n-1)(n^2+n+1)(n^2-n+1)\\ & +n(n^2+n+1)(n^2-n+1)\\ & +(Cn+D)n(n-1)(n^2-n+1)\\ & +(En+F)n(n-1)(n^2+n+1).\end{array}$$This simplifies to $$n^4+4n^3-2n^2-n=(Cn+D)n(n-1)(n^2-n+1)+(En+F)n(n-1)(n^2+n+1).$$Dividing both sides by $n(n-1),$ we get $$-n^2+3n+1=(Cn+D)(n^2-n+1)+(En+F)(n^2+n+1).$$Expanding, we get $$-n^2+3n+1=(C+E)n^3+(C+D+E+F)n^2+(C-D+E+F)n+D+F.$$Matching coefficients, we get $$\begin{array}{rl}C+E& =0,\\ -C+D+E+F& =-1,\\ C-D+E+F& =3,\\ D+F& =1.\end{array}$$Since $C+E=0,$ $-D+F=3.$ Hence, $D=-1$ and $F=2.$ Then $-C+E=-2,$ so $C=1$ and $E=-1.$ Therefore, $$\frac{4n^3-n^2-n+1}{n^6-n^5+n^4-n^3+n^2-n}=\frac{1}{n-1}-\frac{1}{n}+\frac{n-1}{n^2+n+1}-\frac{n-2}{n^2-n+1}.$$Then $$\begin{array}{rl}\sum _{n=2}^{\infty }\frac{4n^3-n^2-n+1}{n^6-n^5+n^4-n^3+n^2-n}& =\left(1-\frac{1}{2}+\frac{1}{7}\right)\\ & +\left(\frac{1}{2}-\frac{1}{3}+\frac{2}{13}-\frac{1}{7}\right)\\ & +\left(\frac{1}{3}-\frac{1}{4}+\frac{3}{21}-\frac{2}{13}\right)+\cdots \\ & =\boxed{1}.\end{array}$$