Chapter 8.2, Problem 56E

To determine

To graph: the curves $y=x^n,0\le x\le 1,\text{for}n=0,1,2,3,\mathrm{4......}$ on a common screen and by finding the area between successive curves, give a geometric demonstration of the fact that ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n(n+1)}}=1$ .

Answer to Problem 56E

Explanation of Solution

Given information: Given curves are $y=x^n,0\le x\le 1,\text{for}n=0,1,2,3,\mathrm{4......}$

Calculation:

The graph of the curves $y=x^n,0\le x\le 1,\text{for}n=0,1,2,3,\mathrm{4......}$ is shown below.

  

So to find the area between every two successive curves using integrals:

  $\begin{array}{l}{A}_{1,2}={\displaystyle \underset{0}{\overset{1}{\int }}(x^0}-x^1)dx\\ A_{3,2}={\displaystyle \underset{0}{\overset{1}{\int }}(x^1}-x^2)dx\\ \text{Doing}\text{this}\text{for}\text{all}\text{curves}\text{.}\\ A_{n.n-1}={\displaystyle \underset{0}{\overset{1}{\int }}(x^n}-x^{n+1})dx\end{array}$

Next to find the sum of all these areas which is the sum of integrals so:

  $\begin{array}{c}{S}_n={\displaystyle {\sum }_{n=0}^{\infty }{\displaystyle \underset{0}{\overset{1}{\int }}(x^n}-x^{n+1})dx}\\ \text{Solve}\text{the}\text{integral}\text{:}\\ S_n={\displaystyle {\sum }_{n=0}^{\infty }{\displaystyle \underset{0}{\overset{1}{\int }}(x^n}-x^{n+1})dx}.\\ ={\displaystyle \sum _{n=0}^{\infty }(\frac{x^{n+1}}{n+1}}-\frac{x^{n+2}}{n+2})|\begin{array}{c}1\\ 0\end{array}\\ ={\displaystyle \sum _{n=0}^{\infty }[}\frac{1^{n+1}}{n+1}-\frac{0^{n+1}}{n+1}-\frac{1^{n+2}}{n+2}+\frac{0^{n+2}}{n+2}]\\ ={\displaystyle \sum _{n=0}^{\infty }[}\frac{1}{n+1}-\frac{1}{n+2}]\\ ={\displaystyle \sum _{n=0}^{\infty }[}\frac{n+2-n-1}{(n+1)(n+2)}].\\ ={\displaystyle \sum _{n=0}^{\infty }[}\frac{1}{(n+1)(n+2)}]\\ \text{Replace}n+1\text{by}n\\ A={\displaystyle \sum _{n=1}^{\infty }[}\frac{1}{n(n+1)}].\end{array}$

The sum of all the areas is the Area of the square with side 1, therefore.

  $A=1$ .

Therefor ${\displaystyle \sum _{n=1}^{\infty }[}\frac{1}{n(n+1)}]=1.$

Hence proved.