Chapter 8.2, Problem 46E

To determine

To find: The sum of the given series using a convenient expression for the partial sum.

Answer to Problem 46E

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^3-n}}=\frac{1}{4}$ .

Explanation of Solution

Given information: Given series is ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^3-n}}$ .

Calculation:

Using the partial fraction command,

  $a=\frac{1}{n^3-n}=\frac{1}{n(n-1)(n+1)}$

Partial sum will be:

  $s_n={\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^3-n}}={\displaystyle \sum _{n=2}^{\infty }\left(\frac{1}{2(n-1)}+\frac{1}{2(n+1)}-\frac{1}{n}\right)}$

Let us find some initial partial sums to observe a pattern,

  $\begin{array}{l}{s}_1=a_2=\frac{1}{2}+\frac{1}{6}-\frac{1}{2}=\frac{1}{6}\\ \\ s_2=a_2+a_3=\frac{1}{6}+\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{3}\right)=\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\\ \\ s_3=a_2+a_3+a_4=\frac{1}{6}+\left(\frac{1}{4}+\frac{1}{8}-\frac{1}{6}\right)+\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{4}\right)=\frac{1}{4}+\frac{1}{10}-\frac{1}{8}\end{array}$

The pattern is,

  $\begin{array}{l}{s}_n=\frac{1}{4}+\frac{1}{2n+4}-\frac{1}{2n+2}\\ \\ \underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\left(\frac{1}{4}+\frac{1}{2n+4}-\frac{1}{2n+2}\right)=\frac{1}{4}\end{array}$

  $\left\{s_n\right\}$ is convergent and its limit is $\frac{1}{4}$ , this implies that the series is convergent and its sum is also $\frac{1}{4}$ .

So, ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^3-n}}=\frac{1}{4}$ .

Using CAS also,

  ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n^3-n}}=\frac{1}{4}$ .

Hence, verified.