+0  
 
+2
45
3
avatar+406 

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

 Oct 1, 2020
 #2
avatar
+2

sumfor(n, 2,1000, (4*n^3-n^2-n+1) / (n^6-n^5+n^4-n^3+n^2-n)) = It converges to 1

 

Partial Sum Formula = (m^3 + m^2 - m - 1)/(m (m^2 + m + 1))

 Oct 1, 2020
 #3
avatar+406 
+1

I get 1, too, and that was correct. Thanks guest smiley

 Oct 1, 2020

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