Chapter 9, Problem 1RWDT

To determine

To find the radius of convergence of the series if power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-2\right)}^n}$ converges conditionally at $x=5.$

Answer to Problem 1RWDT

The possible radius of convergence of ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-2\right)}^n}$ is $3.$

Explanation of Solution

Given:

Power series: ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-2\right)}^n}$ converges conditionally at $x=5.$

Concept Used:

The convergence Theorem for power series: For a power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-c\right)}^n}$ if radius of convergence is $R$ then the series

• Diverges for $\left|x-c\right|>R$ .
• Absolutely converges for $\left|x-c\right|<R\text{ then }$ converges if $\text{ c}-R<x<c+R.$
• The series may or may not converge at either of the endpoints $x=c-R\text{ and }x=c+R.$

Result: If the interval of convergence is $\text{ c}-R<x<c+R,$ conclude that the series diverges (it does not even converge conditionally) for $\left|x-c\right|>R$ , because for those values of $x$ the $n^{th}$ term does not approach zero.

Calculation:

Power series: ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-2\right)}^n}$ converges conditionally at $x=5.$

Compare the given power series with standard power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-c\right)}^n}$ ,

Here, $c=2$ then interval of convergence is $\text{ 2}-R<x<2+R.$

Now, according to the definition of interval of convergence the series is absolutely convergent in the interval of convergence that is, $\text{ 2}-R<x<2+R.$

So, $2+R\le 5$ .

But, according to the result (given above) the series is diverges for $\left|x-2\right|>R\text{ or }x-2>R$ then

  $\left(2+R\right)$ cannot be greater than $5$ . So,

  $\begin{array}{c}2+R=5\\ R=5-2\\ R=3\end{array}$

Hence, the possible radius of convergence of ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-2\right)}^n}$ is $3.$ Also, the radius of convergence of series cannot be greater than $5$ .

Conclusion:

The possible radius of convergence of ${\displaystyle \sum _{n=0}^{\infty }{a}_n{\left(x-2\right)}^n}$ is $3.$