Relationship between order of a pole and decay rate of a taylor series? Out of curiosity, is there a known relationship between the order of a functions pole and the rate of decay (/growth?) of the coefficients of a Taylor Series? Or one that can be proven?
For example, say that we take a taylor series at $z=0$ of a function holomorphic at $z=0$ but with a simple pole at $z=a$. Then the radius of convergence of this taylor series is at most equal to $a$. From my understanding, the terms can also be shown to decay at rate $\dfrac{1}{a^n}$ (so that the series may converge). 
Are there any differences that may occur if the pole is of higher order? Because intuitively, this changes the nature of the function about the pole itself in terms of how quickly it "goes off to infinity" (for lack of a better term) and such. 
 A: Yes, there is a rather straightforward relation. The pole closest to the origin dictates the rate of decay.
First let's look at the function $p_k(z)=\frac{1}{(1-z)^{k}}$. The Taylor series of this function is $\sum_{n\ge0}\binom{n+k-1}{k-1} z^{n}$.
Now, if the pole is at the point $a$, then the function becomes $p_k(z/a)=\frac{1}{(1-z/a)^{k}}$ and obviously the Taylor series becomes $\sum_{n\ge0}\binom{n+k-1}{k-1} a^{-n}z^{n}$.
Now the only thing you need to get your answer is that if $f$ is a meromorphic function we can "substract the poles" and turn it into a holomorphic function. The coefficients of the Taylor series of the resulting holomorphic function decay faster than the ones of $f$.
Edit: let $f$ be meromorphic with a simple pole at $a$, then it is straightforward to check that the function $f(z)-\frac{c}{a-z}$ with the appropriate $c$ is holomorphic.
By the way, you can get a bound of the Taylor coefficients of a holomorphic function if you know its growth rate at infinity by using Cauchy's integral.
