Sum R Cubed Minus R 2E6605

1. **State the problem:** We need to find the sum $$\sum_{r=6}^N \frac{1}{r^3 - r}$$ in terms of $N$. 2. **Rewrite the denominator:** Notice that $$r^3 - r = r(r^2 - 1) = r(r-1)(r+1).$$ So the term becomes $$\frac{1}{r(r-1)(r+1)}.$$ 3. **Partial fraction decomposition:** We want to express $$\frac{1}{r(r-1)(r+1)}$$ as $$\frac{A}{r-1} + \frac{B}{r} + \frac{C}{r+1}.$$ Multiply both sides by $r(r-1)(r+1)$: $$1 = A r (r+1) + B (r-1)(r+1) + C r (r-1).$$ 4. **Expand and collect terms:** $$1 = A (r^2 + r) + B (r^2 - 1) + C (r^2 - r).$$ Group by powers of $r$: $$1 = (A + B + C) r^2 + (A - C) r - B.$$ 5. **Equate coefficients:** - Coefficient of $r^2$: $A + B + C = 0$ - Coefficient of $r$: $A - C = 0$ - Constant term: $-B = 1 \implies B = -1$ 6. **Solve the system:** From $A - C = 0$, we get $A = C$. From $A + B + C = 0$, substitute $B = -1$ and $C = A$: $$A - 1 + A = 0 \implies 2A = 1 \implies A = \frac{1}{2}.$$ Then $C = A = \frac{1}{2}$. 7. **Rewrite the term:** $$\frac{1}{r(r-1)(r+1)} = \frac{1/2}{r-1} - \frac{1}{r} + \frac{1/2}{r+1}.$$ 8. **Sum from $r=6$ to $N$:** $$\sum_{r=6}^N \frac{1}{r^3 - r} = \sum_{r=6}^N \left( \frac{1/2}{r-1} - \frac{1}{r} + \frac{1/2}{r+1} \right).$$ 9. **Separate sums:** $$= \frac{1}{2} \sum_{r=6}^N \frac{1}{r-1} - \sum_{r=6}^N \frac{1}{r} + \frac{1}{2} \sum_{r=6}^N \frac{1}{r+1}.$$ 10. **Rewrite indices:** $$\sum_{r=6}^N \frac{1}{r-1} = \sum_{k=5}^{N-1} \frac{1}{k},$$ $$\sum_{r=6}^N \frac{1}{r} = \sum_{k=6}^N \frac{1}{k},$$ $$\sum_{r=6}^N \frac{1}{r+1} = \sum_{k=7}^{N+1} \frac{1}{k}.$$ 11. **Combine:** $$= \frac{1}{2} \sum_{k=5}^{N-1} \frac{1}{k} - \sum_{k=6}^N \frac{1}{k} + \frac{1}{2} \sum_{k=7}^{N+1} \frac{1}{k}.$$ 12. **Express sums using harmonic numbers $H_m = \sum_{i=1}^m \frac{1}{i}$:** $$\sum_{k=5}^{N-1} \frac{1}{k} = H_{N-1} - H_4,$$ $$\sum_{k=6}^N \frac{1}{k} = H_N - H_5,$$ $$\sum_{k=7}^{N+1} \frac{1}{k} = H_{N+1} - H_6.$$ 13. **Substitute:** $$= \frac{1}{2} (H_{N-1} - H_4) - (H_N - H_5) + \frac{1}{2} (H_{N+1} - H_6).$$ 14. **Simplify:** $$= \frac{1}{2} H_{N-1} - \frac{1}{2} H_4 - H_N + H_5 + \frac{1}{2} H_{N+1} - \frac{1}{2} H_6.$$ 15. **Group harmonic numbers:** $$= \left( \frac{1}{2} H_{N-1} - H_N + \frac{1}{2} H_{N+1} \right) + \left( -\frac{1}{2} H_4 + H_5 - \frac{1}{2} H_6 \right).$$ 16. **Final answer:** $$\sum_{r=6}^N \frac{1}{r^3 - r} = \frac{1}{2} H_{N-1} - H_N + \frac{1}{2} H_{N+1} - \frac{1}{2} H_4 + H_5 - \frac{1}{2} H_6,$$ where $H_m = \sum_{i=1}^m \frac{1}{i}$ is the $m$-th harmonic number.