# Comples analysis - Radius of convergence of a Taylor series question

1. May 17, 2012

1. The problem statement, all variables and given/known data
Find the radius of convergence of the Taylor series at z = 1 of the function:

$\frac{1}{e^{z}-1}$

2. Relevant equations

3. The attempt at a solution
Hi everyone,

Here's what I've done so far.

Multiply top and bottom by minus 1 to get:

-1/(1-e^z)

And then this is the expansion of an infinite geometric series:

1/(1-z) = Ʃ(z^n)

Then use the formula 1/R = lim sup (nth root of absolute value of a_n)

where here a_n is constant, 1.

So then you will get that the lim sup of this is 1.

So then R = 1.

However, I don't like this solution as I think that geometric formula only applies if |z|<1, which is not the case for e^z.
I am a bit stuck as to how else to write out the Taylor series though. And how do I make use of the fact z=1? Does that mean I have to write out a Taylor series in the form Ʃa_n.(z-1)^n?

I'd really appreciate if someone could please point me in the right direction.

Thanks

2. May 17, 2012

### Citan Uzuki

You want to find the radius of convergence of the series of the form $\sum_{n=0}^{\infty} a_n (z-1)^n$, yes. Don't bother looking at the series itself though. Just use the fact that the radius of convergence of a Taylor series is just the distance from its center to the nearest singularity. Where are the singularities of $\frac{1}{e^z - 1}$?

3. May 17, 2012

Ah. So the singularity would be at z=0 then?
So the distance from z=0 to z=1 would be 1, so the radius of convergence is then R=1 centered at the point z=1?

4. May 17, 2012

Also, just out of interest, how exactly would you write that formula as a power/Laurent series?

5. May 17, 2012

### Citan Uzuki

That's the closest singularity, yes. But for future reference, remember that there are also other singularities at $2\pi ik$ for every $k\in \mathbb{Z}$.

Yep. You got it!

It's easy enough to find the Taylor series of $e^z - 1$ directly, and then you can just use long division to get the terms of the Taylor series for $\frac{1}{e^z-1}$. It's a pretty tedious calculation though.

6. May 17, 2012

### scurty

If you restrict values of z so that $\left|\frac{1}{e^z}\right| < 1$,

$\begin{eqnarray*} \displaystyle \frac{1}{e^z - 1} &=& \left(\frac{1}{e^z}\right)\left(\frac{1}{1-\frac{1}{e^z}}\right)\\ &=& \left(\frac{1}{e^z}\right) \cdot \sum_{n=0}^{\infty} \left(\frac{1}{e^z}\right)^n\\ &=& \sum_{n=0}^{\infty} \left(\frac{1}{e^z}\right)^{n+1}\\ &=& \sum_{n=0}^{\infty} (e^z)^{-n-1}\\ &=& \sum_{n=-\infty}^{0} (e^z)^{n-1}\\ &=& \sum_{n=-\infty}^{-1} (e^z)^n \end{eqnarray*}$

Edit: I just realized this isn't in power series form, you need a $(z-a)^n$.. I'll leave this here because it's a good technique anyway. Yeah, it would be tough to put it into power series form.

7. May 18, 2012