+0

+2
87
3
+475

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

Oct 1, 2020

#2
+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
+475
+1

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

Oct 1, 2020