Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$

Question

Im trying to find the radius of convergence for the taylor sum of the function $\frac{z}{e^z-1}$ around z=0.

So far I've found the coffiecients $a_0=1$, $a_1=-\frac{1}{2}$, $a_2=\frac{1}{12}$ and that $a_{2k+1}=0$ for $k\geq 1$. It's pretty ugly to look for the rest of the terms in the series.

Is there a neat way to find that?

No, I guess what i mean is that we expand this function around 0, when $f(0)=\lim_{z\rightarrow 0}f(z)$ — Snufsan, Jan 07 '15 at 22:34
$0$ is clearly a false singularity of this meromorphic function on $\mathbb{C}$. It can obviously be extended as an holomorphic function on a neighborhood of $0$ by setting $f(0)=1$. — Philippe Malot, Jan 07 '15 at 22:39
since $e^{2i\pi} = 1,$ would not the radius of convergence $< 2 \pi?$ — abel, Jan 07 '15 at 22:45

Philippe Malot · Answer 1 · 2015-01-08T11:43:03.907

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Hint: The radius of convergence of a series at $z_0$ is the distance from $z_0$ to (one of) its closest non-removable singularity.

edited Jan 08 '15 at 11:43

answered Jan 07 '15 at 22:44

Philippe Malot

2,667

Can you solve the equation $e^z-1=0$? – Philippe Malot Jan 07 '15 at 22:47
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... closest non-removable singularity. In general they need not be poles. – Robert Israel Jan 07 '15 at 23:24
Good remark! I'll modify my answer. – Philippe Malot Jan 08 '15 at 11:42

Finding radius of convergence for taylor series of $\frac{z}{e^z-1}$

1 Answers1