Compute \[\frac{1}{2^3 - 2} + \frac{1}{3^3 - 3} + \frac{1}{4^3 - 4} + \dots + \frac{1}{100^3 - 100}.\]

Guest Dec 6, 2020

#1**0 **

Hello Guest! This problem i was looking at and it is actually quite easy, this relates to the sigma notation. here we go:

First we can find the rule, this is $1/(n^3-n)$, next simplifying the rule it is $1/(n(n^2-1))$, using the conjugate thing we can find the rule if $1/((n-1)(n)(n+1))$

Next, we must try to solve and expand, we can make an expression

$1/2(1/(n(n-1))-1/(n(n+1))$, we expand and make the sigma notation rule. After solving I put my work in latex and here it is:

\begin{align*}

\sum_{n = 2}^\infty \frac{1}{n^3 - n} &= \sum_{n = 2}^\infty \left( \frac{1/2}{n - 1} - \frac{1}{n} + \frac{1/2}{n + 1} \right) \\

&= \left( \frac{1/2}{1} - \frac{1}{2} + \frac{1/2}{3} \right) + \left( \frac{1/2}{2} - \frac{1}{3} + \frac{1/2}{4} \right) + \left( \frac{1/2}{3} - \frac{1}{4} + \frac{1/2}{5} \right) \\

&\quad + \dots + \left( \frac{1/2}{98} - \frac{1}{99} + \frac{1/2}{100} \right) + \left( \frac{1/2}{99} - \frac{1}{100} + \frac{1/2}{101} \right) \\

&= \frac{1/2}{1} - \frac{1/2}{2} - \frac{1/2}{100} + \frac{1/2}{101} \\

&= \boxed{\frac{5049}{20200}}.

\end{align*}

Guest Dec 7, 2020