### Instruction: Compute
\[\sum_{n = 2}^\infty \frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n}.\].### Response: I notice that the numerator and denominator of the fraction are both polynomials of degree 6, so I suspect that the fraction will simplify as n gets large.
I also notice that the coefficients of the numerator and denominator alternate between positive and negative, so I wonder if there is some factorization that will cancel out some terms.
To simplify the fraction, I divide both the numerator and denominator by the highest power of n, which is n^6.
This gives me
\[\frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n} = \frac{4 - \frac{1}{n} - \frac{1}{n^2} + \frac{1}{n^3}}{1 - \frac{1}{n} + \frac{1}{n^2} - \frac{1}{n^3} + \frac{1}{n^4} - \frac{1}{n^5}}.\]
Now I see that as n gets larger and larger, the terms that involve fractions of n will get smaller and smaller, and approach zero.
So the fraction will approach 4 as n approaches infinity.
This means that the terms of the series will also approach 4 as n approaches infinity, and the series will not converge, since the terms do not get closer to zero.