Chapter 8.2, Problem 8E

To determine

To find: at least 10 partial sums of the series .Graph both the sequence of terms and the sequence of partial sums on the same screen and found does it appear that the series is convergent or divergent. If it is convergent, find the sum. If it is divergent, explain.

Answer to Problem 8E

This series is convergent and its limit is $\frac{5}{12}$ .

Explanation of Solution

Given information: Given series is ${\displaystyle \sum _{n=2}^{\infty }\frac{1}{n(n+2)}}$

Calculation:

In this case, general term and nth partial sum are:

  $\begin{array}{l}{a}_n=\frac{1}{n(n+2)}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+2})\\ S_n={\displaystyle \sum _{i=2}^{n+1}{a}_i}\frac{1}{2}(\frac{1}{i}-\frac{1}{i+2})\end{array}$

First find two or three partial sum in order to try to find some pattern for the rest of them:

  $\begin{array}{l}{S}_1=a_2=\frac{1}{2}(\frac{1}{2}-\frac{1}{4})=\frac{1}{8}\\ S_2=a_2+a_3=\frac{1}{8}+\frac{1}{2}(\frac{1}{3}-\frac{1}{5})=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5})=\frac{23}{120}\\ S_3=a_2+a_3+a_4=\frac{1}{2}((\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6}))=\frac{1}{2}((\frac{1}{2}+\frac{1}{3}-\frac{1}{5}-\frac{1}{6})=\frac{7}{30}\end{array}$

So, can conclude that the pattern is:

  $S_n=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3})$ .

Get the following terms:

  $\begin{array}{l}{S}_4=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}-\frac{1}{7})=\frac{11}{42}.\\ S_5=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{7}-\frac{1}{8})=\frac{95}{336}\\ S_6=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{8}-\frac{1}{9})=\frac{43}{144}\\ S_7=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{9}-\frac{1}{10})=\frac{14}{45}\\ S_8=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{10}-\frac{1}{11})=\frac{53}{165}\\ S_9=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{11}-\frac{1}{12})=\frac{29}{88}\\ S_{10}=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{12}-\frac{1}{13})=\frac{35}{104}\\ \end{array}$

The limit of a partial sum is the following :

  $\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-\frac{1}{n+2}-\frac{1}{n+3})=\frac{1}{2}(\frac{1}{2}+\frac{1}{3}-0-0)=\frac{5}{12}.$

The conclusion is that this series is convergent and its limit is $\frac{5}{12}$ .

The graph of the sequences of terms and partial sum is given below.