Chapter 9, Problem 70RE

$\text{(a)}$

To determine

To prove: The statement ${\displaystyle \sum _{n=1}^{\infty }\frac{n(n+1)}{x^n}}=\frac{2x^2}{{(x-1)}^3}$

Explanation of Solution

Given information:

For $\text{|x|>1}$ by differentiating the identity ${\displaystyle \sum _{n=1}^{\infty }{x}^{n+1}}=\frac{x^2}{1-x}$ twice, multiplying the result by $x$ and then replacing $x$ by $\frac{1}{x}$ .

Calculation:

Differentiate the equation first. By utilize the Quotient Rule and Term-by-Term Differentiation,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }(n+1)}{x}^n=\frac{(2x)(1-x)-\left(x^2\right)(-1)}{{(1-x)}^2}\\ {\displaystyle \sum _{n=1}^{\infty }(n+1)}{x}^n=\frac{2x-x^2}{{(1-x)}^2}\end{array}$

Differentiate the equation first. By utilize the Quotient Rule and Term-by-Term Differentiation,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }n}(n+1)x^{n-1}=\frac{(2-2x){(1-x)}^2-\left(2x-x^2\right)(2(1-x)(-1))}{{(1-x)}^4}\\ {\displaystyle \sum _{n=1}^{\infty }n}(n+1)x^{n-1}=\frac{2{(1-x)}^3+2x(2-x)(1-x)}{{(1-x)}^4}\\ {\displaystyle \sum _{n=1}^{\infty }n}(n+1)x^{n-1}=\frac{2}{{(1-x)}^3}\end{array}$

By multiplying the result of the previous step by $x$ , obtain,

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

Replace

  $x$ by $\frac{1}{x}$ .

Then,

  $\begin{array}{c}{\displaystyle \sum _{n=1}^{\infty }n}(n+1){\left(\frac{1}{x}\right)}^n=\frac{2\cdot \frac{1}{x}}{{\left(1-\frac{1}{x}\right)}^3}\\ {\displaystyle \sum _{n=1}^{\infty }n}(n+1)\frac{1}{x^n}=\frac{\frac{2}{x}}{{\left(\frac{x-1}{x}\right)}^3}\\ {\displaystyle \sum _{n=1}^{\infty }\frac{n(n+1)}{x^n}}=\frac{2}{x}\cdot \frac{x^3}{{(x-1)}^3}\\ {\displaystyle \sum _{n=1}^{\infty }\frac{n(n+1)}{x^n}}=\frac{2x^2}{{(x-1)}^3}\end{array}$

Therefore, the required statement ${\displaystyle \sum _{n=1}^{\infty }\frac{n(n+1)}{x^n}}=\frac{2x^2}{{(x-1)}^3}$

  $($ proved $)$ .

$\text{(b)}$

To determine

To find: The real solution greater than $1$ of the equation.

Answer to Problem 70RE

The answer: $x\approx 2.77$

Explanation of Solution

Given information:

The equation is $x={\displaystyle \sum _{n=1}^{\infty }\frac{n(n+1)}{x^n}}$

Calculation:

From part $\text{(a)}$ :

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

Since the given equation equivalent to,

  $x=\frac{2x^2}{{(x-1)}^3}$

When, $\text{|x|>1}$ thus the solution of this equation is also a solution of the original equation.

Then,

  ${(x-1)}^3=2x$

To use the graphing calculator,

  $x\approx 2.77$

Therefore, the required real solution greater than $1$ of the equation is $x\approx 2.77$ .