Sum

Compute

$\mathbf{\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10}}$

$\begin{array}{|rcll|} \hline s&=& \dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10} \\\\ &=& \dfrac{1}{2(2^2-1)} + \dfrac{1}{3(3^2-1)} + \dfrac{1}{4(4^2-1)} + \dots + \dfrac{1}{10(10^2-19} \\\\ &=& \dfrac{1}{(2-1)2(2+1)} + \dfrac{1}{(3-1)3(3+1)} + \dfrac{1}{(4-1)4(4+1)} + \dots + \dfrac{1}{(10-1)10(10+1)} + \dots + \dfrac{1}{(n-1)n(n+1)} \\\\ \hline s &=&\sum \limits_{n=2}^{10}\dfrac{1}{(n-1)n(n+1)} \\ \\ &&\boxed{ \frac{1}{(n-1)n} =\frac{1}{n-1}-\frac{1}{n} \\ \frac{1}{(n+1)n} =\frac{1}{n}-\frac{1}{n+1}\\\ldots\\ \frac{1}{(n-1)n(n+1)} = \frac12\left( \frac{1}{n-1}-\frac{2}{n} +\frac{1}{n+1} \right) } \\\\ s &=&\dfrac12\sum \limits_{n=2}^{10}\left( \dfrac{1}{n-1}-\dfrac{2}{n} +\dfrac{1}{n+1} \right) \\ \\ &=&\dfrac12\left( \sum \limits_{n=2}^{10}\dfrac{1}{n-1}-\sum \limits_{n=2}^{10}\dfrac{2}{n} +\sum \limits_{n=2}^{10}\dfrac{1}{n+1} \right) \\ \\ &=&\dfrac12\left( \sum \limits_{n=1}^{9}\dfrac{1}{n} -\sum \limits_{n=2}^{10}\dfrac{2}{n} +\sum \limits_{n=3}^{11}\dfrac{1}{n} \right) \\ \\ &&\boxed{ \sum \limits_{n=1}^{9}\frac{1}{n} = \frac11 + \frac12 + \sum \limits_{n=3}^{9}\frac{1}{n} \\ -\sum \limits_{n=2}^{10}\frac{2}{n} = -\frac{2}{2}-\frac{2}{10}-2\sum \limits_{n=3}^{9}\frac{1}{n} \\ \sum \limits_{n=3}^{11}\frac{1}{n} = \frac{1}{10}+\frac{1}{11}+\sum \limits_{n=3}^{9}\frac{1}{n} } \\\\ s &=&\dfrac12\left( \dfrac11 + \dfrac12 + \sum \limits_{n=3}^{9}\dfrac{1}{n} -\dfrac{2}{2}-\dfrac{2}{10}-2\sum \limits_{n=3}^{9}\dfrac{1}{n} +\dfrac{1}{10}+\dfrac{1}{11}+\sum \limits_{n=3}^{9}\dfrac{1}{n}\right) \\\\ &=&\dfrac12\left( \dfrac11 + \dfrac12 -\dfrac{2}{2}-\dfrac{2}{10}+\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac12\left( \dfrac12 -\dfrac{2}{10}+\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac12\left( \dfrac12 -\dfrac{1}{10}+\dfrac{1}{11}\right) \\ \\ &=&\dfrac14 -\dfrac{1}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{5}{20} -\dfrac{1}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{4}{20}+\dfrac{1}{22} \\ \\ &=&\dfrac{1}{5}+\dfrac{1}{22} \\ \\ &=&\dfrac{22+5}{5*22}\\ \\ \mathbf{s}&=&\mathbf{\dfrac{27}{110}}\\ \hline \end{array}$

$\dfrac{1}{2^3 - 2} + \dfrac{1}{3^3 - 3} + \dfrac{1}{4^3 - 4} + \dots + \dfrac{1}{10^3 - 10} = \mathbf{\dfrac{27}{110}}$