Let $F_0(z) := \frac{z}{e^z-1}$ for $z\in\mathbb{C}\setminus \{2\pi\mathrm{i}\mathbb{Z}\}\bigcup\{0\}$.
Define the sequences of Bernoulli numbers $(B_n)_{n\in\mathbb{N}_0}$ in the neighbourhood of $0$ through $$F_0(z) = \sum_{n=0}^\infty B_n\frac{z^n}{n!}$$
Show $$\sum_{k=0}^{n-1}\binom{n}{k}B_k = 0$$ for $n\gt1$ and $B_0 = 1$ with the help of the relation: $(e^z - 1)F_0 = z$.
Furthermore, show $B_{2k+1} = 0$ for $k\gt0$.
Ok, I can see that $F_0$ is not holomorph for $z_0\in\{2\pi\mathrm{i}\mathbb{Z}\}\bigcup\{0\}$. To remove the isolated singularity in $z_0$ we use the Riemann's theorem, which says:
Let $D\subset\mathbb{C}$ be an open subset of the complex plane, $a\in D$ a point of $D$ and $f$ a holomorphic function defined on the set $D\setminus\{a\}$. The following are equivalent:
- $f$ is holomorphically extendable over $a$
- $f$ is continuously extendable over $a$
- There exists a neighborhood of $a$ on which $f$ is bounded.
- $\lim \limits_{z \to a}(z-a)f(z)=0$
So if I use 4. for our example it would be: $\lim \limits_{z \to 0}(z-0)F_0(z)=\lim \limits_{z \to 0}(z-0)\frac{z}{e^z-1}\overset{!}{=}0$.
The limit is $0$, because $(z-0)\xrightarrow{z\rightarrow0}0$ and $\lim \limits_{z \to 0}\frac{z}{e^z-1}$=1. I just showed that the Riemann's theorem is fulfilled, so I know there exist a funktion $\widetilde{f}|D\setminus {z_0}=f$ which is holomorph on whole $D$.
But it doesn't help me much and I don't know the next step. Is $F_0(z)=\sum_{n=0}^\infty B_n\frac{z^n}{n!}$ our $\widetilde{f}$? It is defined at $z_0$. How can I use $\widetilde{f}$ to solve this excercise?
First: $\widetilde{f} = F_0$ (the quotient) in a perforated neighborhood of $0$, so $\widetilde{f} = F_0$ (the series) in a neighborhood of $0$. For the formula: $$z = F_0(z)(e^z - 1) = \left(\sum_{n=0}^\infty B_n\frac{z^n}{n!}\right) \left(\sum_{n=1}^\infty \frac{z^n}{n!}\right) = $$ $$ = \sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}B_k\frac{z^k}{k!}\frac{z^{n -k}}{(n - k)!}\right) = \sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}\frac{n!}{k!(n - k)!}B_k\right)\frac{z^n}{n!}. $$ Finally, you can check easily that $$z\longmapsto F_0(z) + \frac{z}2 = \frac{z(e^z + 1)}{2(e^z - 1)}$$ is even.