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Determine \(\sum_{n = 2}^\infty \frac{4n^3 - n^2 - n + 1}{n^6 - n^5 + n^4 - n^3 + n^2 - n}.\)

 Feb 13, 2020
 #1
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By partial fractions, the sum works out to 2.

 Feb 13, 2020
 #2
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∑[(4 n^3 - n^2 - n + 1)/(n^6 - n^5 + n^4 - n^3 + n^2 - n)], n, 2, ∞ =converges to 1

 Feb 13, 2020

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