Let $f:\mathbb{C}\to\mathbb{C}$ be defined as:
$$f(z)=\frac{1}{e^z-1}-\frac{1}{z}+\frac 12$$
I need to prove that there exists a sequence $\{B_k\}_{k=1}^{\infty}$ of complex numbers, such that $f(z)$ is given by a series of the form:
$$f(z)=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}B_k}{(2k)!}z^{2k-1}$$
And to conclude that $\limsup_\limits{k\to\infty}{\sqrt[k]{B_k}}=\infty$.
First, I proved that $\lim_\limits{z\to0}f(z)=0$, so if I define $f(0)=0$, then the new $f(z)$ would be continuous. However, it doesn't mean that $f(z)$ is analytic. So how do I know that I can expand $f(z)$ to a Taylor Series? (I want to expand $f(z)$ to a Taylor Series since the given series looks a lot like one, with $f^{(2k)}(0)=(-1)^{k-1}B_k$ for every $k\in\mathbb{N}$. This would eventually be an additional thing that I need to prove, somehow, don't know yet).
Another thing I noticed is that $f(z)$ is an odd function, so if $f(z)$ has a Taylor Expansion with $a_k$ as coefficients, then $a_{2k}=0$ for every $k\in\mathbb{N}$ (This correlates with what I need to prove).
Finally, I know that to get to the conclusion, I just need to prove that the radius of convergence of the series would be $R=0$, and then just use Cauchy Hadamard Theorem.
So on the one hand, I have a feeling I know what to do; But on the other it seems that I'm missing something. Thanks!
The function $f$ is not defined on $\mathbb C$, but on $\mathbb C \setminus 2 \pi i \mathbb Z$. In each $z_k = 2k \pi i$ it has an isolated singularity. For $k = 0$ the singularity is removable since $\lim_{z \to 0} f(z) = 0$, for all other $k$ it is not removable. Thus $f$ has a homolomorphic extension to $(\mathbb C \setminus 2 \pi i \mathbb Z) \cup \{0\}$ and it can be expanded into a Taylor series around $0$. The radius of convergence of this series is $2\pi$.
Since $f$ is odd, we a priori know that the Taylor series has the form of your question with suitable $B_k$. Now assume that $\limsup_ {k\to \infty} \sqrt[k]{\lvert B_k \rvert} < \infty$. Hence there exists $C < \infty$ such that $\sqrt[k]{\lvert B_k \rvert} < C$ for almost all $k$. Then $$\limsup_ {k\to \infty} \sqrt[k]{\left\lvert \dfrac{(-1)^{k-1} B_k}{(2k)!} \right\rvert} \le \limsup_ {k\to \infty} \dfrac{C}{\sqrt[k]{(2k)!}} = 0 .$$ This would imply that the radius of convergence of your series is $\infty$ which is a contradiction because it is $2\pi$.
Remark: For the radius of convergence of the Taylor series see Radius of convergence of Taylor series of holomorphic function which shows that it is $\ge 2\pi$. By the identity theorem for holomorphic functions it cannot be $> 2\pi$ because $f$ has a non-removable singularity on the boundary of the disk with radius $2\pi$ around $0$.