Showing the radius of convergence for a power series is equal to the radius of convergence for its derivative

Question

Consider the power series:

$$ \sum_{n=0}^{\infty} a_n (x - c)^n $$

Now consider its derivative:

$$ \sum_{n=1}^{\infty} n a_n (x - c)^{n-1} $$

We can say at first that the Radius of Convergence for the original power series is

$$ R = \lim_{n \to \infty} |a_{n+1} / a_{n}| $$

(via the Ratio Test).

On the other hand, can we not also say that the radius of convergence for the derivative of the power series is

$$ \lim_{n \to \infty} \left|\frac{(n+1) a_{n+1}}{n a_{n}} \right| = |a_{n+1} / a_{n}| = R? $$

via the same argument? Is my reasoning correct? That is, is the argument that the Radius of Convergence the same for both a power series and its derivative really this simple? :)

Answer 1

Observe that

$$\lim\sup_{n\to\infty}\sqrt[n]{|a_n|}=\lim\sup_{n\to\infty}\sqrt[n]{|na_n|}$$

since $\;\sqrt[n]n\xrightarrow[n\to\infty]{}1\;$ , so both power series convergence radius are the same.

Answer 2

Suppose the radius of convergence of $\sum a_nx^n$ is $R.$ Then $\sum a_nx^n$ converges absolutely for $x\in (-R,R).$ Now fix an $x_0 \in (-R,R),$ and choose $y\in (-R,R)$ with $|y| > |x_0|.$ Then $n|a_nx_0^n| = |a_ny^n|n|x_0/y|^n.$ Because $|x_0/y| < 1,$ $n|x_0/y|^n \to 0.$ Since $\sum |a_ny^n| < \infty,$ $\sum |na_nx_0^n| < \infty.$ It follows that the radius of convergence of $\sum na_nx^n$ is at least $R.$