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? 🙂
Best Answer
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.