Chapter 10.5, Problem 48E

To determine

The interval of convergence of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(4x-5\right)}^{2n+1}}{n^{3/2}}}$

Answer to Problem 48E

The interval of convergence and the series converges absolutely is $\left[1,\frac{3}{2}\right]$

Explanation of Solution

Given information:

The given series is

  ${\displaystyle \sum _{n=1}^{\infty }\frac{{\left(4x-5\right)}^{2n+1}}{n^{3/2}}}$

Formula used:

The ratio test is used.

Calculation:

The series of absolute values is

  ${\displaystyle \sum _{n=1}^{\infty }\left|a_n\right|}={\displaystyle \sum _{n=1}^{\infty }\frac{{\left|4x-5\right|}^{2n+1}}{n^{3/2}}}$

The ratio Test let ${\displaystyle \sum a_n}$ be a series with positive terms, and with

  $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=L$

Then, the series converges if L < I, the series diverges if L >1 and the test is inconclusive if L=1

Using the ratio test, we check for absolute convergence as follows,

  $\begin{array}{l}\underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\underset{n\to \infty }{\lim }\frac{{\left|4x-5\right|}^{2n+1}}{{\left(n+1\right)}^{3/2}}\times \frac{n^{3/2}}{{\left|4x-5\right|}^{2n+1}}\\ ={\left|4x-5\right|}^2\underset{n\to \infty }{\lim }\left(\frac{n}{n+1}\right)\\ \underset{n\to \infty }{\lim }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}={\left|4x-5\right|}^2\end{array}$

The series converges absolutely for

  $\left|4x-5\right|<1\text{or}x\in \left(1,\frac{3}{2}\right)$

And diverges for

  $\left|4x-5\right|>1$

And when | x | = 1, the series is

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

This converges absolutely by the p-test

And when $x=\frac{3}{2}$ , the series is

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

This converges absolutely by the p-test

(a) Interval of convergence: $\left[1,\frac{3}{2}\right]$

(b) Series converges absolutely on $\left[1,\frac{3}{2}\right]$

(c) None

Conclusion:

The interval of convergence and the series converges absolutely is $\left[1,\frac{3}{2}\right]$