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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 $lim_{n \rightarrow \infty}$ $a_n/a_{n+1} = z_0.$

It is clear that the radius of convergence is $|z_0|$ and so that if it does converge, it converges to something with distance from the origin $|z_0|.$ But the rest I am not sure about. Some help would be fantastic. Thank you.

kingkongdonutguy
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  • Let $\displaystyle c={\rm Res}(f,z_0)\not =0$ the residue of $f(z)$ at $z_0$, and put $\displaystyle g(z)=f(z)-\frac{c}{z-z_0}$. – Kelenner Jul 22 '14 at 14:45

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Since $f$ has a simple pole in $z_0$, $$\lim_{z\to z_0}(z-z_0)\,f(z) = C, $$ hence by replacing $f$ with its Taylor series in zero, $$\lim_{z\to z_0}\left(-a_0 z_0 + \sum_{n=0}^{+\infty} (a_n-z_0 a_{n+1})\, z^{n+1}\right)=C.$$ This gives that $\{(a_n-z_0 a_{n+1})\,z_0^{n+1}\}_{n\in\mathbb{N}}$ is a Cauchy sequence, from which: $$\forall \varepsilon>0,\forall n\geq N,\quad |a_n-z_0 a_{n+1}|\leq \frac{\varepsilon}{|z_0|^{n+1}}$$ follows. Hence: $$\left|\frac{a_n}{a_{n+1}}-z_0\right|\leq\frac{\varepsilon}{|a_{n+1}|\cdot|z_0|^{n+1}},$$ but given that $f(z)=g(z)+\frac{C}{z-z_0}$ where $g(z)$ is an entire function, we have: $$ |a_n| \geq \frac{C}{|z_0|^n}-2\pi\frac{\sup_{|z|=|z_0|} |g(z)|}{n!|z_0|^n}$$ by the Liouville inequality. So: $$\left|\frac{a_n}{a_{n+1}}-z_0\right|\leq\frac{\varepsilon}{C-\frac{2\pi}{(n+1)!}\sup_{|z|=|z_0|} |g(z)|}$$ and: $$\lim_{n\to +\infty}\frac{a_{n}}{a_{n+1}}=z_0$$ as wanted.

Jack D'Aurizio
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  • I don't get how you took care of the rhs – kingkongdonutguy Jul 22 '14 at 15:20
  • You need a lower bound for $|a_{n+1}|\cdot|z_0|^{n+1}$ that you can grasp by writing $a_{n+1}$ through the Cauchy formula and assuming without loss of generality that $|f(z)|\geq M$ as soon as $z$ is close to $z_0$. – Jack D'Aurizio Jul 22 '14 at 15:29
  • But $f$ is not analytic at $z = z_0$. I'm confused – kingkongdonutguy Jul 22 '14 at 15:36
  • Estimate $a_n$ by integrating $f(z)$ in a circle around the origin that passes near $z_0$. – Jack D'Aurizio Jul 22 '14 at 15:38
  • But then $|f(z)|$ is not necessarily $\geq M$ except at a couple points. Unless you do a small circle near $z_0$. But then can't say anything about $a_{n + 1}$. – kingkongdonutguy Jul 22 '14 at 15:48
  • Also I don't see how any of this helps, because we need to know the sign of $f(z)$. – kingkongdonutguy Jul 22 '14 at 15:50
  • The sign of a complex-valued function?! You just need to know that since $z_0$ is a simple pole for $f(z)$, the values of $f(z)$ in a punctured neighbourhood of $z_0$ belong to a neighbourhood of infinity. – Jack D'Aurizio Jul 22 '14 at 16:09
  • Or, just to make things clearer, write $f(x)=g(x)+\frac{C}{z-z_0}$ where $g(z)$ is an entire function. You have that $a_n$ behaves like $C/z_0^n$ plus a negligible perturbation given by the coefficients of an entire function. – Jack D'Aurizio Jul 22 '14 at 16:11
  • I'm trying to calculate $2\pi ia_{n + 1} =$ $\int_{C_{z_0 - \epsilon}(0)}$$\frac{f(z)}{z^{n + 2}}dz$ where ${C_{z_0 - \epsilon}(0)}$ is a circle at 0 almost touching $z_0$. – kingkongdonutguy Jul 22 '14 at 16:12
  • I changed the last part of the proof to make it clearer. – Jack D'Aurizio Jul 22 '14 at 16:18