Chapter 8, Problem 3RCC

(a)

To determine

To find: The geometric series and the under what circumstances it is convergent and find the sum.

Answer to Problem 3RCC

Thegeometric series is defined and geometric series converges to $\left|r\right|<1$ and for $\left|r\right|>1$ the series diverges.

Explanation of Solution

Calculation:

Consider that the geometric series are of the sum $a+ar+a{r}^2+\mathrm{.......}a{r}^{n-1}+\mathrm{...}={\displaystyle \sum _{n=1}^{\infty }a{r}^{n-1}}$ . Here, $a$ and $r$ is the fixed real numbers and $a\ne 0$ the series is written as ${\displaystyle \sum _{n=0}^{\infty }a{r}^n}$ .

If $\left|r\right|<1$ the geometric series is of the form $a+ar+a{r}^2+\mathrm{......}a{r}^{n-1}+\mathrm{..}$ converges to $\left[\frac{a}{1-r}\right]$ .

Therefore, ${\displaystyle \sum _{n=1}^{\infty }a{r}^{n-1}}=\frac{a}{1-r}$ . Here, $\left|r\right|<1$ and for $\left|r\right|>1$ , the series diverges.

(b)

To determine

To find: The p series and what circumstances it converges.

Answer to Problem 3RCC

The series of the form $\frac{1}{1^p}+\frac{1}{2^p}+\frac{1}{3^p}+\mathrm{....}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ is p series and the series is convergent for $p>1$ and diverges for $p\le 1$ .

Explanation of Solution

Calculation:

Consider that the $p$ series is shown below,

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

Here, $p>0$ is the p series.

The series is convergent for $p>1$ and diverges for $p\le 1$ .

Thus, the series of the form $\frac{1}{1^p}+\frac{1}{2^p}+\frac{1}{3^p}+\mathrm{....}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^p}}$ is p series and the series is convergent for $p>1$ and diverges for $p\le 1$ .