Chapter 8.2, Problem 41E

To determine

To find: The value of x for ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{3^n}}$is convergent.

Answer to Problem 41E

The sum is $\frac{x}{3-x}$

Explanation of Solution

Given information:

The geometric series ${\displaystyle \sum _{n=1}^{\infty }a{r}^{n-1}}$

  $=a+ar+a{r}^2+\mathrm{....}+a{r}^{n-1}$

  $=a[1+r+r^2+\mathrm{....}+r^{n-1}]$is convergent for $\left|r\right|<1$

Calculation:

Now ,

  ${\displaystyle \sum _{n=1}^{\infty }\frac{x^n}{3^n}}=\frac{x}{3}+\frac{x^{}}{3^2}+\frac{x^3}{3^3}+\mathrm{....}+\frac{x^n}{3^n}$

  $=\frac{x}{3}[1+\left(\frac{x}{3}\right)+{\left(\frac{x}{3}\right)}^2+{\left(\frac{x}{3}\right)}^3+\mathrm{....}+\frac{x}{3}]$

is convergent for $\left|\frac{x}{3}\right|<1$ ………(1) i.e $\left|x\right|<3$

Hence , value of x, $-3<x<3$

To , find the sum :

We know that the sum of geometric series is $=\frac{a}{1-r}$

From (1) clearly $a=\frac{x}{3}$

and $r=\frac{x}{3}$

then the sum is $=\frac{\frac{x}{3}}{1-\frac{x}{3}}$

  $=\frac{x}{3}\times \frac{3}{3-x}$

  $=\frac{x}{3-x}$

  $$