Let $f(z)=\sum_{k \geq 0} a_k z^k$ be a power series with radius of convergence $R >0$ on the disk $D(0,R)$.

Show that if the series $\sum_{k \geq 0} f^{(k)}(0)$ converges, then $R = \infty$.

To be honest, I do not know where to start. I know the derivative of $f(z)$, but I'm stuck to show the main result.

Could anyone give me a good hint to solve the problem?

  • $\begingroup$ What is the relation between the coefficients $a_k$ and the derivatives $f^{(k)}(0)$? $\endgroup$ – Daniel Fischer Feb 14 '16 at 21:42
  • $\begingroup$ Both series will converge to zero when $k \to \infty$... This is what you mean? $\endgroup$ – user1050421 Feb 14 '16 at 21:48
  • $\begingroup$ No, there is a very specific relation between the coefficients of the Taylor series of a function and the derivatives of that function at the centre of the expansion. $\endgroup$ – Daniel Fischer Feb 14 '16 at 21:52
  • $\begingroup$ I'm not very aged; I don't know very well the taylor series. Would you tell me more or simply refer to a website that I can read and understand myself. $\endgroup$ – user1050421 Feb 14 '16 at 21:55
  • 1
    $\begingroup$ You can start with the wikipedia article. $\endgroup$ – Daniel Fischer Feb 14 '16 at 21:59

The radius of convergence is given by $$ \frac1R=\limsup_{n\to\infty}\sqrt[n]{|a_n|}$$ Assume $R<\infty$. Then there exists $c>0$ (namely any positive $c<\frac 1R$) such that there are infinitely many $n$ with $\sqrt[n]{|a_n|}>c$, i.e., $|a_n|>c^n$. One quickly verifies that $f^{(n)}(0)=n!a_n$. Hence in the series $\sum f^ {(n)}(0)$ there are infinitely many summands $>n!c^n$. For $n>\frac 2{c^2}$, at least $\frac n2$ of the factors making $n!$ are $>\frac1{c^2}$, hence for such $n$, $|f^{(n)}(0)|>\left(\frac1{c^2}\right)^{n/2}c^n=1$, contradicting convergence of $\sum f^{(n)}(0)$.

  • $\begingroup$ From the "For $n \dots$", could you explain what you did? I know stategically what you try to do (i.e. $\lim_{n \to \infty} f^{(n)}(0) \not= 0$). If you can just explain in more details what you did from this line, it should be very kind. $\endgroup$ – user1050421 Feb 14 '16 at 23:07

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