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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!

Amit Zach
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  • Using the more precise definition of $f$ (with $f(0)=0$), you can just verify the Cauchy-Riemann equations in order to obtain that $f$ is holomorphic and thus also analytic – Maximilian Janisch Jan 05 '20 at 14:41
  • You can also just apply Riemann's Theorem on removable singularities to conclude that $f$ is analytic in a neighbourhood of $0$. – Bruno Krams Jan 05 '20 at 17:25
  • By the way: Since the function is analytic around the origin, it's radius of convergence is positive! – Bruno Krams Jan 05 '20 at 17:27
  • You should prove that the composition of two analytic functions (on the corresponding domains) is analytic, using that the two power series converge absolutely thus we can change the order of summation. Since $1/s$ is analytic at $s=1$ then $\frac1{\frac{e^z-1}{z}}$ is analytic at $z=0$, this proves that $zf(z)$ is analytic and since it has a zero at $0$.. – reuns Jan 05 '20 at 18:21

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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 $0$.

Paul Frost
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