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I am studying for a complex qual and I am having a little difficulty with the following problem. Any help that you could give me would be awesome. Thank you.

Let $f(z) = \sum_{n=0}^\infty a_nz^n$ be the Taylor series around $0$ of a function which is analytic in $\mathbb{C}$ \ ${z_0},$ $z_0\neq 0$ and has only a simple pole at $z_0$. Prove that $lim_{n\rightarrow \infty} a_n/a_{n+1} = z_0.$

kingkongdonutguy
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  • The key is probably the fact that $$g(z)=f(z)(z-z_0)=(z-z_0)\sum_{n=0}^\infty a_nz^n= a_0z_0+\sum_{n=0}^\infty (a_{n-1}-z_0a_n)z^n$$ is entire. – N. S. Jun 10 '14 at 02:00
  • Is the term $a_{n-1}$ correct? –  Mar 18 '16 at 01:46

2 Answers2

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I decided to add another answer rather than delete or change the last since the first answer is nice ill leave it, but here is a complete solution, independent of previous considerations. $(z-z_0)f(z)$ is an entire function with power expansion $$\sum (a_n -a_{n+1}z_0)z^{n+1}$$ now as this series converges for all $z$, and the original series has a finite radius of convergence we must have $$\lim |\frac{a_n -a_{n+1}z_0}{a_{n+1}}|=0$$

And this is exactly what was to be proved.

Rene Schipperus
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This is only a partial answer, but a power series converges on the largest radius where the function is defined. The series cannot converge at this point $z_0$. Therefore the radius of convergence is $|z_0|$ now by the ratio test this gives,

$$\lim\limits_{n\to \infty} \frac{|a_n|}{|a_{n+1}|}=|z_0|$$

Of course I have not used that $z_0$ is a simple pole so somehow that must be involved in proving the better limit.

Rene Schipperus
  • 39,526