Finding the Laurent Series for $\frac{1}{e^z-1}$ for $0<|z|<2\pi$

Question

Since $\left|\dfrac{1}{e^z}\right|<1$ I figured I could rewrite the given function into a geometric series:

$$\sum_{n=1}^{\infty} \frac{1}{(e^z)^n}$$

But this seems to be way off the mark. I think I am confused about when one is supposed to rewrite the expressions into series. I guess this expression above isn't correct because I need to express $e^z$ into a series. I can't really make sense of how to do that though.

Should I use $\displaystyle e^z=1+\frac{z}{1!}+\frac{z^2}{2!}+...=\sum_{n=0}^{\infty}\frac{z^n}{n!}$

giving

$$\frac{1}{e^z-1}=\sum_{n=1}^{\infty} \frac{1}{(\sum_{j=0}^{\infty}\frac{z^j}{j!})^n}$$

This doesn't make a lot of sense to me.

Answer 1

One may observe that $\displaystyle z \to \frac{z}{e^z-1}$ is analytic in $0<|z|<2\pi$, then it admits a power series expansion $$ \frac{z}{e^z-1}=\sum\limits_{n=0}^{\infty}b_n\frac{z^n}{n!}, \quad 0<|z|<2\pi, \tag1 $$ with $b_0=1$, $b_1=-1/2$.

Then, multiplying $(1)$ by $\displaystyle e^z$, $$ \left(\sum\limits_{k=0}^{\infty}b_k\frac{z^k}{k!}\right)\left(\sum\limits_{m=0}^{\infty}\frac{z^m}{m!}\right)=\frac{z}{e^z-1}e^z=z+\frac{z}{e^z-1}=z+\sum\limits_{n=0}^{\infty}b_n\frac{z^n}{n!},\tag2 $$ using the Cauchy product, one gets $$ b_n=\sum_{k=0}^n \binom{n}{k}b_k, \quad n>1, \quad b_0=1,\, b_1=-1/2,\tag3 $$ but from $(3)$ one sees that these are just the standard Bernoulli numbers: $B_n$.

Finally, we have

$$ \frac1{e^z-1}=\frac1z-\frac12+\frac{z}{12}+\cdots=\frac1z+\sum\limits_{n=0}^{\infty}B_{n+1}\frac{z^n}{(n+1)!}, \quad 0<|z|<2\pi. \tag4 $$

Answer 2

Note: If you proceed with the geometric series approach, you find (formally) that:

$$\frac{1}{e^z-1}=\sum_{n\ge1}e^{-nz}=\sum_{n\ge1}\sum_{k\ge0}\frac{(-1)^kn^kz^k}{k!}=\sum_{k\ge0}\frac{(-1)^kz^k}{k!}\sum_{n\ge1}n^k$$

The term at the end, $\sum_{n\ge1}n^k$, is obviously infinite and divergent. That being said, some treatments allow us to assign value to these sums in some sense. By noting the definition of the Riemann zeta function, we can write $\sum_{n\ge1}n^k=\zeta(-k)$.

What is somewhat surprising is that if we compute the coefficients of the expansion for $\frac{1}{e^z-1}$ and compare them to these divergent sums, they agree with the results obtained by analytic continuation of the Riemann zeta function, that is, that all this haphazard symbolic manipulation turns out to be (at least partially) vindicated.

A cute consequence of this is that because $\frac{1}{e^z-1}$ is (nearly) an odd function, the "value" of $\sum_{n\ge1}n^k$ is $=0$ for all even $k\ge2$.