Homework Help: 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