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Compute $\frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \frac{1}{4^3 - 4} + \dots + \frac{1}{6^3 - 6}.$

$\phantom{100^3 - 100}$

Guest Jun 27, 2022

Guest · Answer 1 · 2022-06-27T05:21:02

sumfor(n, 2, 6, 1 / (n^3 - n)==It converges to 5 / 21