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I have the function $ f(z)=\frac{z}{e^z-1} $. I'm looking for a power series.

My idea:

I define a power series $$ \frac{z}{e^z-1}=f(z):=\sum\limits_{k=0}^\infty c_k\cdot z^k. $$ Now I start equating the coefficients $ c_k $:

$$ \begin{align}z&=(e^z-1)\cdot \left(\sum\limits_{k=0}^\infty c_k\cdot z^k \right)=e^z\left( \sum\limits_{k=0}^\infty c_k\cdot z^k\right)-\sum\limits_{k=0}^\infty c_k\cdot z^k\\&=\left(\sum\limits_{k=0}^\infty \frac{1}{k!}\cdot z^k\right)\cdot \left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)\\&=\sum\limits_{k=0}^\infty\left(\sum\limits_{l=0}^k \frac{1}{l!}\cdot z^l\cdot c_{k-l}\cdot z^{k-l}\right)-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right) \\&=\sum\limits_{k=0}^\infty\left(\sum\limits_{l=0}^k \frac{1}{l!}\cdot c_{k-l}\cdot z^{k}\right)-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)\\&=\sum\limits_{k=0}^\infty\left(\sum\limits_{l=0}^k \frac{1}{l!}\cdot c_{k-l}\right)\cdot z^{k}-\left(\sum\limits_{k=0}^\infty c_k\cdot z^k\right)\\&=\sum_{k=0}^\infty \left[\left(\sum\limits_{l=0}^k\frac{1}{l!}\cdot c_{k-l} \right)-c_k \right]\cdot z^k\\&=\sum_{k=0}^\infty \left(\sum\limits_{l=1}^k\frac{1}{l!}\cdot c_{k-l} \right)\cdot z^k\end{align}$$

$ \deg k=0: \quad 0=\sum\limits_{l=1}^0\frac{1}{l!}\cdot c_{k-l}=0\\\deg k=1: \quad 1=\sum\limits_{l=1}^1\frac{1}{l!}\cdot c_{k-l}=c_0\\\deg k\geq 2:\quad 0=\sum\limits_{l=1}^k\frac{1}{l!}\cdot c_{k-l}=\left(\sum\limits_{l=2}^k\frac{1}{l!}\cdot c_{k-l}\right) +c_{k-1}\\\qquad \qquad \ \ c_{k-1}=-\sum\limits_{l=2}^k\frac{1}{l!}\cdot c_{k-l}$.

I struggle to show that $c_{k-1}$ converges absolutely (I tried ratio test and root test) and I'm not to find an explizit formula of $c_k$ and taylor approximation is a catastrophy concerning of the derivatives of $ f $. How else could I show that this series converges absolutely?

FFjet
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hallo007
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    This series is famous as the (exponential) generating function of the Bernoulli numbers: $t/(e^t - 1) = \sum_{n \geq 0} (B_n/n!)t^n$, where $B_n$ is the $n$th Bernoulli number. It is hopeless to expect a tidy formula for the coefficients. As a function of a complex variable the first pole is at $t = 2\pi i$, so the radius of convergence of this series is $R = 2\pi$. I am not sure what you mean by "show that this series converges absolutely": at which $t$ were you trying to show that? It converges absolutely for $|t| < 2\pi$ and not for $|t| > 2\pi$. – KCd Dec 21 '20 at 03:08
  • I wanted to show that at $ t=0. $ At first I belived that this series converges for all $ t $ which is not the case here. – hallo007 Dec 21 '20 at 03:22
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    It would be better in that case to write the function as $1/((e^t-1)/t)$ because $e^t - 1 = t + t^2/2! + t^3/3! + \cdots$ for all $t$ implies $(e^t-1)/t = 1 + t/2! + t^2/3! + \cdots$ for all $t$, including at $t = 0$. That leaves as your remaining task showing that if $g(t)$ is a power series in $t$ with a positive radius of convergence and $g(0) \not= 0$, the reciprocal function $1/g(t)$ can be written on an interval around $0$ (or disc, if working in $\mathbf C$) a power series with a positive radius of convergence. This is much easier to do with complex analysis than real analysis. – KCd Dec 21 '20 at 03:26
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    It is not easy to predict the radius of convergence of an unknown power series on the real line without investigating its behavior in the complex plane. It is as a function on $\mathbf C$, not $\mathbf R$, that you can often find a clear understanding of what the radius of convergence is, especially when the function is a ratio of functions with well-understood power series. The simple example of $1/(x^2+1)$ illustrates the point: on $\mathbf R$ this function is infinitely differentiable everywhere, but on $\mathbf C$ it blows up at $x = \pm i$. That is "why" its radius of convergence is $1$. – KCd Dec 21 '20 at 03:29

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This is the generating function for the Bernoulli numbers:

https://en.wikipedia.org/wiki/Bernoulli_number

marty cohen
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