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

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? :)

• I changed the title. Sep 4, 2016 at 16:32
• The radius of convergence is not always given by the ratio test. When the ratio test does give it, however, your argument works.
– zhw.
Sep 4, 2016 at 17:05
• @zhw: Do you mean that the ratio test only gives the radius of convergence AFTER you have shown that the power series does indeed converge? Sep 4, 2016 at 17:06
• No, I'm saying what I wrote. For example $x + x^2/2^2 + x^3 + x^4/2^4 + x^5 + x^6/2^6 + \cdots$ has radius of convergence 1, but the ratio test fails miserably here.
– zhw.
Sep 4, 2016 at 17:10

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.

• What if this limsup equals $1$? By the root test, this may be inconclusive. Sep 4, 2016 at 17:35
• @user1770201 You seem to have missed the point. Google "Cauchy-Hadamard formula" or something like that: it gives you the radius of convergence $\;R\;$ by means of the formula $$R=\lim\sup_{n\to\infty}\frac1{\sqrt[n]{|a_n|}}$$ and thus what my answer proves is that both a power series and its derivative power series have the same radius of convergnece. Sep 4, 2016 at 17:46
• @Anu you could use, I guess, that $$\limsup_{n\to\infty}\sqrt[n]n=\lim_{n\to\infty}\sqrt[n]n=1\ldots$$ Feb 4, 2018 at 8:46
• @F.Tomas$$|a_n|^{\frac1{n-1}}=\left(|a_n|^{1/n}\right)^{\frac n{n-1}}$$ Jul 26, 2020 at 14:13
• @DonAntonio, even if $|a_n|^{\frac{1}{n-1}}=(|a_n|^{1/n})^{\frac{n}{n-1}}$ how if follows that $\limsup |a_n|^{\frac{1}{n}}=\limsup |a_n|^{\frac{1}{n-1}}$? I am bit confused.
– RFZ
Jun 23, 2021 at 23:33

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 $R.$

• This is wrong. The coefficients of x^n is not na_n Feb 24, 2019 at 14:46
• @JohnMitchell I don't understand your comment. I showed that for each $x_0\in (-R,R),$ the series $\sum na_nx_0^n$ converges absolutely. This implies the ROC of $\sum na_nx^n$ is at least $R.$
– zhw.
Feb 24, 2019 at 17:06
• No what you've shown is that $\sum na_nx^n$ converges. We've to show $\sum (n+1)a_(n+1)x^n$ converges Mar 4, 2019 at 18:32
• @JohnMitchell That follows trivially from what I showed.
– zhw.
Mar 4, 2019 at 19:10
• Why do you pause here? Are there examples where the derivative $f'$ of a power series $f$ has a larger convergence disc than $f$? Jan 1 at 1:20