Chapter 6, Problem 253RE

True or False? In the following exercises, justify your answer with a proof or a counterexample.

253. If the radius of convergence for a power series ${\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{a}_n{x}^n}$ is 5, then the radius of convergence for the series ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}n{a}_n{x}^{n-1}}$ is also 5.

To determine

To check: The given statement whether it is true or false and also justified it.

Answer to Problem 253RE

The statement is true.

Explanation of Solution

Given information:

The given statement is:

The radius of convergence for a power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ is 5, then the radius of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }n{a}_n{x}^{n-1}}$ is also 5.

Concept used:

The radius of convergence R for a power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ is given by,

  $R=\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|$ .

Calculation:

The radius of convergence for a power series ${\displaystyle \sum _{n=0}^{\infty }{a}_n{x}^n}$ is given by,

  $\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|=5\left(\text{Given}\right)$

Now, the radius of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }n{a}_n{x}^{n-1}}$ is given by,

  $\begin{array}{c}R=\underset{n\to \infty }{\lim }\left|\frac{\left(n+1\right)a_{n+1}}{n{a}_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\left(1+\frac{1}{n}\right)\frac{a_{n+1}}{a_n}\right|\\ =\underset{n\to \infty }{\lim }\left|\frac{a_{n+1}}{a_n}\right|\\ =5\end{array}$

Therefore, the radius of convergence for the series ${\displaystyle \sum _{n=1}^{\infty }n{a}_n{x}^{n-1}}$ is also 5.

Conclusion:

Hence, the given statement is true.