Computing the residue of $\frac{1}{(1-e^{-z})^n}$ at $z=0$

Question

The question is to compute $$\operatorname{res}_0 \frac{1}{(1-e^{-z})^n}$$ for natural number $n$.

I’ve tried directly calculate the integral by definition and Laurent expansion, but not making any progress. The Laurent series of the function become very nasty for large $n$. Is there any other way to solve this? Any help is much appreciated.

Answer 1

To find the residue, we can integrate $\frac{1}{(1-e^{-z})^{n}}$ around a rectangular contour with vertices at $z=\pm R \pm i \pi$, $R \ge 1$.

Doing so, we get $$\int_{-R}^{R} \frac{1}{\left(1-e^{-(t- i \pi)}\right)^{n}} \, dt + \int_{-\pi}^{\pi} \frac{1}{\left(1-e^{-(R+ it)}\right)^{n}} \, i \, dt - \int_{-R}^{R} \frac{1}{\left(1-e^{-(t+ i \pi)}\right)^{n}} \, dt$$ $$- \int_{-\pi}^{\pi} \frac{1}{\left( 1-e^{-(-R + it)}\right)^{n}} \, i \, dt = 2 \pi i \operatorname{Res}\left[\frac{1}{(1-e^{-z})^{n}},0 \right].$$

Since $e^{-t+i \pi} = -e^{-t} = e^{-t-i \pi}$, the first and third integrals cancel each other.

And as $R \to \infty$, the fourth integral vanishes since $$\left|\int_{-\pi}^{\pi} \frac{1}{\left( 1-e^{-(-R + it)}\right)^{n}} \, i \, dt \right|\le \int_{-\pi}^{\pi} \frac{dt}{\left(e^{R}-1\right)^{n}} = \frac{2 \pi}{(e^{R}-1)^{n}}. $$

(Due to symmetry, it appears that the value of integral might actually be zero for all values of $R >0$.)

That leaves us with $$\operatorname{Res}\left[\frac{1}{(1-e^{-z})^{n}},0 \right] = \lim_{R \to \infty} \frac{1}{2 \pi i} \int_{-\pi}^{\pi} \frac{1}{\left(1-e^{-(R+ it)}\right)^{n}} \, i \, dt. $$

But since $$\left| \frac{i}{\left(1-e^{-(R+it)}\right)^{n}}\right| \le \frac{1}{(1-e^{-R})^{n}} \le 2^{n}$$ for $R \ge 1$, the dominated convergence theorem permits us to move the limit inside the integral and conclude that $$\operatorname{Res}\left[\frac{1}{(1-e^{-z})^{n}},0 \right] = \frac{1}{2\pi i} \int_{-\pi}^{\pi} \, i\, dt = 1.$$

Answer 2

(Too long to be a comment)

Starting with the series $$\frac{z}{1-\mathrm{e}^{-z}}=\sum_{m=0}^\infty B_m^+ \frac{z^m}{m!}.$$

we get $$\frac{1}{(1-\mathrm{e}^{-z})^n}=\left[ \frac{1}{z} \sum_{m=0}^\infty B_m^+ \frac{z^m}{m!} \right]^n=\frac{1}{z^n} \sum_{m=0}^\infty C_m z^m $$ where $$C_m=\sum_{m_{n-1}=0}^{m} \cdots \sum_{m_2=0}^{m_3} \sum_{m_1=0}^{m_2} \frac{B_{m_1}^+}{m_1!} \frac{B_{m_2-m_1}^+}{(m_2-m_1)!} \cdots \frac{B_{m-m_{n-1}}^+}{(m-m_{n-1})!}$$ Numerical evidence suggest that the residue is identically 1, hence from a combinatorical perspective we want to show that

$$\sum_{m_{n-1}=0}^{n-1} \cdots \sum_{m_2=0}^{m_3} \sum_{m_1=0}^{m_2} \frac{B_{m_1}^+}{m_1!} \frac{B_{m_2-m_1}^+}{(m_2-m_1)!} \cdots \frac{B_{n-1-m_{n-1}}^+}{(n-1-m_{n-1})!} \equiv 1$$ for all natural $n$. Perhaps you could show this by induction, using the recursive definition of the Bernoulli numbers.