+0

# Sum

0
47
1

Compute $\frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \frac{1}{4^3 - 4} + \dots + \frac{1}{10^3 - 10}.$

May 15, 2022

#1
+9457
0

You can just use a calculator, but here's a calculation trick:

The denominator looks like $$n^3 - n = n(n - 1)(n + 1)$$ we can perform partial fractions on $$\dfrac1{n(n-1)(n+1)}$$ to get $$\dfrac12 \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12 \left(\dfrac1n - \dfrac1{n + 1}\right)$$. The original expression is $$\displaystyle \sum_{n=2}^{10}\left(\dfrac12 \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12\left(\dfrac1n - \dfrac1{n + 1}\right)\right)=\dfrac12\displaystyle \sum_{n=2}^{10} \left(\dfrac1{n - 1} - \dfrac1n\right) - \dfrac12\displaystyle \sum_{n=2}^{10} \left(\dfrac1n - \dfrac1{n + 1}\right)$$. Now note that each sum is a telescoping sum, so the answer is $$\dfrac12 \left(1 - \dfrac1{10}\right) - \dfrac12 \left(\dfrac12 - \dfrac1{11}\right)$$. You can simplify it by hand or by calculator.

May 15, 2022