7

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? :)

user1770201
  • 5,361
  • 6
  • 44
  • 80
  • I changed the title. – user1770201 Sep 04 '16 at 16:32
  • 1
    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 04 '16 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? – user1770201 Sep 04 '16 at 17:06
  • 2
    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 04 '16 at 17:10

2 Answers2

13

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.

DonAntonio
  • 214,715
  • What if this limsup equals $1$? By the root test, this may be inconclusive. – user1770201 Sep 04 '16 at 17:35
  • 2
    @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. – DonAntonio Sep 04 '16 at 17:46
  • @DonAntonio I was looking for a proof of the fact that $\limsup_{n \rightarrow \infty} |a_n|^\frac{1}{n} = \limsup_{n \rightarrow \infty} |na_n|^\frac{1}{n} $. What is the standard way of doing it? – Anu Feb 04 '18 at 06:39
  • 2
    @Anu you could use, I guess, that $$\limsup_{n\to\infty}\sqrt[n]n=\lim_{n\to\infty}\sqrt[n]n=1\ldots$$ – DonAntonio Feb 04 '18 at 08:46
  • Wouldn't the terms in the derivative series be shifted? How would we show that $$\limsup_{a_n \rightarrow \infty} |a_n|^{\frac{1}{n}} = \limsup_{a_n \rightarrow \infty} |a_n|^{\frac{1}{n-1}}?$$ – F.Tomas Jul 26 '20 at 12:40
  • 1
    @F.Tomas$$|a_n|^{\frac1{n-1}}=\left(|a_n|^{1/n}\right)^{\frac n{n-1}}$$ – DonAntonio Jul 26 '20 at 14:13
  • 1
    @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 '21 at 23:33
  • @ZFR Lim sup of $;|a_n|^{1/n};$ is the usual way to calculate the convergence radius (its reciprocal, in fact), and $;\frac n{n-1}\xrightarrow[n\to\infty]{}1;$ ... – DonAntonio Jun 24 '21 at 14:27
  • @DonAntonio, Please sorry but I still did not get your point. Could you give some details, please? – RFZ Jun 24 '21 at 14:30
  • Since $$\limsup_{n\to\infty}\sqrt[n]{|a_n|}=\frac1R$$ then also $$\limsup_{n\to\infty}\left(\sqrt[n]{|a_n|}\right)^{\frac n{n-1}}=\frac1R$$ You could use here the squeeze theorem, in case the limit is finite. And if the limit is infinite then it will be infinite in the other case as well. – DonAntonio Jun 26 '21 at 18:03
  • @user1770201 There's nothing inconclusive about that. It just means $R=1.$ – Allawonder Jan 01 '24 at 01:23
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 $R.$

zhw.
  • 107,943