Chapter 8.3, Problem 5E

To determine

To distinguish: Between the series ${\displaystyle \sum _{n=1}^{\infty }{n}^b}$ and ${\displaystyle \sum _{n=1}^{\infty }{b}^n}$ .

Answer to Problem 5E

  ${\displaystyle \sum _{n=1}^{\infty }{n}^b}$ is a p-series which converges for $\left|b\right|>1$ and ${\displaystyle \sum _{n=1}^{\infty }{b}^n}$ is a geometric series which converges for $\left|b\right|<1$ .

Explanation of Solution

Given information: Given series are ${\displaystyle \sum _{n=1}^{\infty }{n}^b}$ and ${\displaystyle \sum _{n=1}^{\infty }{b}^n}$ .

Consider the series ${\displaystyle \sum _{n=1}^{\infty }{n}^b}$ .

Let $b<0$ .

Then ${\displaystyle \sum _{n=1}^{\infty }{n}^b}={\displaystyle \sum _{n=1}^{\infty }{n}^{-\left|b\right|}={\displaystyle \sum _{n=1}^{\infty }\frac{1}{n^{\left|b\right|}}}}$

This is a p-series which is convergent if $\left|b\right|>1$ and divergent if $\left|b\right|\le 1$ .

Now, consider the series ${\displaystyle \sum _{n=1}^{\infty }{b}^n}$ .

It is a geometric series which is convergent if $\left|b\right|<1$ and divergent if $\left|b\right|\ge 1$ .

Thus, ${\displaystyle \sum _{n=1}^{\infty }{n}^b}$ is a p-series which converges for $\left|b\right|>1$ and ${\displaystyle \sum _{n=1}^{\infty }{b}^n}$ is a geometric series which converges for $\left|b\right|<1$ .