About the limit of the coefficient ratio for a power series over complex numbers

Solution 1:

Hint: Assume that you have a simple pole at $z = z_0$, where $|z_0| = 1$ and try to prove it. In particular, take $f(z) = \frac{g(z)}{z-z_0}$ where $g(z)$ is holomorphic on $\Omega$. Prove the result for this case. (Expand $\frac{1}{z-z_0}$ about $z=0$ and do some manipulations). Now the same idea can be extended for higher order poles.

EDIT: For a simple pole, $f(z) = \frac{g(z)}{z-z_0} = \displaystyle \sum_{n=0}^{\infty} a_n z^n$. Since $g(z)$ is holomorphic, $g(z) = \displaystyle \sum_{n=0}^{\infty} b_n z^n$. So $\displaystyle \sum_{n=0}^{\infty} b_n z^n = (z-z_0) \displaystyle \sum_{n=0}^{\infty} a_n z^n \Rightarrow b_{n+1} = a_n - z_0 a_{n+1}$.

Now what can we say about $\displaystyle \lim_{n \rightarrow \infty} b_n$ and $\displaystyle \lim_{n \rightarrow \infty} a_n$?

(Note: $g(z)$ holomorphic on $\Omega$ whereas $f(z)$ is holomorphic except at $z_0$, a point on the unit disc).

This same idea will work for higher order poles as well.

Solution 2:

Let's try another way to solve this problem.

Construct a contour, which consists of two parts. The first part is a circle and is a little bit larger than the unit circle, except near the point $z_0$. We call it $C_1$, and make it have absolute value strictly larger than $1+\delta$ for some $delta$. The second part is a small circle with radius $\epsilon$.

Suppose $z_0$ as pole have degree k, then $f(\zeta)=\frac{g(z)}{(z-z_0)^k}$, where $g(z)$ is holomorhpic.

For $C_1$, we have that $|\int_{C_1}\frac{g(\zeta)}{\zeta^n}\frac{1}{(\zeta-z_0)^k}d\zeta|\leq \frac{1}{\epsilon^k}\int_{C_1}\frac{M}{(1+\delta)^n}d\zeta \to 0$

For $C_\epsilon$, we have $$\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+1}}\frac{1}{(\zeta-z_0)^k}d\zeta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+1}}e^{-i\theta k} d\theta$$

In the same way,

$$\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+2}}\frac{1}{(\zeta-z_0)^k}d\zeta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+2}}e^{-i\theta k} d\theta$$

By multiplying the second one with $z_0$, and computing the difference, we get,

$$ \Delta=\int_{-\theta_0}^{-\pi+\theta_0} \frac{g(z_0+\epsilon e^\theta_0)}{(z_0+\epsilon e^\theta_0)^{n+2}}e^{-i\theta k} \epsilon e^{i\theta}d\theta \to 0$$, as $\epsilon \to 0$

which means

$$ \frac{\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+1}}\frac{1}{(\zeta-z_0)^k}d\zeta}{z_0\int_{C_\epsilon}\frac{g(\zeta)}{\zeta^{n+2}}\frac{1}{(\zeta-z_0)^k}d\zeta} \to 1 $$ as $\epsilon \to 0$

Combining all of these and Cauchy's integral formuals that $a_0=f(0)=\frac{1}{2\pi i}\int_C\frac{f(\zeta)}{\zeta}d\zeta$ and $n!a_n=f^{(n)}(0)=\frac{n!}{2\pi i}\int_C\frac{f(\zeta)}{\zeta^{n+1}}d\zeta$, we split $C$ as $C_1$ and $C_\epsilon$, we onle need to choose carefully the $\epsilon$'s and $\delta$'s to complete our proof.