Chapter 10.1, Problem 53E

To determine

To show: The particular result for the given function.

Explanation of Solution

Given information:

The series given

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

Calculation:

The series given is ${\displaystyle \sum }_{n=1}^{\infty }a{r}^{n-1}$ using the notation v $S_n=a+ar+a{r}^2+a{r}^3+\dots \dots \dots +a{r}^{n-1}$

Since, $S_n=\frac{a-a{r}^n}{1-r}$

If $\left|r\right|<1,$ , then

  $\underset{n\to \infty }{\lim }{r}^n=0$

Therefore,

  ${\displaystyle \sum }_{n=1}^{\infty }a{r}^{n-1}=\underset{n\to \infty }{\lim }{S}_n=\underset{n\to \infty }{\lim }\frac{a-a{r}^n}{1-r}=\frac{a}{1-r}$

  ${\displaystyle \sum }_{n=1}^{\infty }a{r}^{n-1}=\frac{a}{1-r},$

  $\left|r\right|<1$

If $\left|r\right|>1$ or $r=-1,$ , then $r^n$ has no finite limit as $n\to \infty ,$ , so the expression $\frac{a-a{r}^n}{1-r}$ has no finite limit. Therefore, ${\displaystyle \sum }_{n=1}^{\infty }a{r}^{n-1}$ diverges.

If $r=1$ , then $n^{\text{,"}}$ partial is na which goes to $\pm \infty$