Laurent-series expansion of $1/(e^z-1)$

Solution 1:

Let $f(z) = (e^z - 1)^{-1}$ which has a simple pole at $z = 0$ (easy enough to see). Consider $$ h(z) = \frac{e^z - 1}{z} = \sum_{n = 0}^{\infty}\frac{z^n}{(n + 1)!} $$ $h$ is an entire function (prove it to yourself). Now let $g(z) = \frac{1}{h(z)}$ which is analytic over some area (I leave where as an exercise). Now $$ f(z) = \frac{1}{zh(z)} = \frac{g(z)}{z} = \sum_{n = -\infty}^{\infty}a_nz^n $$ Furthermore, $a_n$ can be found \begin{align} a_n &= \frac{1}{2\pi i}\int_{|z| = R}\frac{f(z)}{z^{n+1}}dz\\ &= \frac{1}{2\pi i}\int_{|z| = R}\frac{g(z)}{z^{n+2}}dz \end{align} where $0<R<2\pi$. $g$ is analytic on the inside of $|z| = R$. By Cauchy's Theorem, $a_n = 0$ for $n\leq -2$. $$ f(z) = \sum_{k = -1}^{\infty}a_nz^n $$ Now compute the first few $a_n$. To find the derivative of $g$, we should first find the derivative of $h$. $$ h^{(k)}(z) = \sum_{n=k}^{\infty}\frac{n!z^{n-k}}{(n-k)!(n+1)!} $$ Therefore, $$ h^{(k)}(0) = \frac{1}{k+1} $$ for all $k\geq 0$ We can easily see that $1 = g(z)h(z)$ so $0=g'h+gh'$. In general, $$ 0 = (gh)^{(k)}(z) = \sum_{i = 0}^k\binom{k}{i}h^{k-i}(z)g^i(z) $$ At $z = 0$, $h^{k-i}(0) = \frac{1}{k - i + 1}$; therefore, $$ 0 = (gh)^{(k)}(0) = \sum_{i = 0}^k\binom{k + 1}{i}g^i(0) $$ Going back to the coefficient $a_n$, we have $$ a_n = \frac{1}{2\pi i}\int_{|z| = R}\frac{g(z)}{z^{n+2}}dz = \frac{g^{(n+1)}(0)}{(n+1)!} $$ for all $n\geq -1$. $$ 0 = \sum_{j = 0}^k\frac{a_{j-1}}{(k-(j-1))!} $$ So $a_{-1} = 1$, $a_0 = -1/2$, all positive even terms are zero.... Let $B_k := (-1)^{k-1}(2k)!a_{2k-1}$ be Bernoulli numbers. Note that $F(z) = \frac{1}{e^z - 1} -\frac{1}{z} + \frac{1}{2}$ is an odd function. Therefore, $$ f(z) = \frac{1}{z} -\frac{1}{2} +\sum_{k=1}^{\infty}a_{2k-1}z^{2k-1} = \frac{1}{z} -\frac{1}{2} +\sum_{k=1}^{\infty}(-1)^{k-1}\frac{B_k}{(2k)!}z^{2k-1} $$

Solution 2:

The other answer has its benefits, but if you want to check your answer or simply bash out the "important" coefficients and ignore all that insignicantly small stuff, then here goes. Just note that we need to be careful about convergence.

We have that $\frac{1}{e^z-1} = \frac{1}{z(1+(z/2!+z^2/3!+...))}$ where we let $(z/2!+z^2/3!+...) = P(z)$. Then:

$$\frac{1}{e^z-1} = 1/z - P(z)/z + P(z)^2/z - P(z)^3/z + ... \\ = 1/z - 1/2 - z/3! + z/(2!)^2 - z^2/4! + 2z^2/(2!\cdot 3!) - z^2/(2!)^3 + O(z^3) \\ = 1/z - 1/2 + z/12 + z^2\cdot (1/6-1/24-1/8) + O(z^3) \\ = 1/z - 1/2 + z/12 + O(z^3).$$

Solution 3:

(This is a rewrite of dustin's answer for my own reference.)

Clearly $f(z)=\frac{1}{e^z - 1}$ has a pole at $z=0$. Since $\lim_{z\to0}zf(z)=1$, the pole is simple. Thus $f(z)$ has a Laurent series expansion $\sum_{n\ge-1}a_nz^n$ about zero with $a_{-1}=1$. Now, as both $$g(z)=\frac{z}{e^z - 1}=\sum_{n\ge0}a_{n-1}z^n$$ and $$\frac{1}{g(z)}=\frac{e^z - 1}{z}=\sum_{n\ge0}\frac{z^n}{(n+1)!}$$ are analytic at zero, we have $$\left(\sum_{n\ge0}a_{n-1}z^n\right)\left(\sum_{n\ge0}\frac{z^n}{(n+1)!}\right)=1.$$ By comparing coefficients on both sides, we see that $a_0=-\frac12$ and $\{a_n\}_{n\ge-1}$ is given by the recurrence relation $a_{-1}=1$ and $$\sum_{k=0}^n\frac{a_{k-1}}{(n-k+1)!}=0\quad(n\ge1).$$ In particular, $a_0=-\frac12$. Finally, in a deleted neighbourhood of zero, it is straightforward to verify that $$\sum_{n\ge1}a_nz^n=f(z)-\frac{a_{-1}}{z}-a_0 =\frac{1}{e^z-1}-\frac{1}{z}+\frac{1}{2}$$ is an odd function. Therefore, all positive even terms $a_2,a_4,\ldots$ are actually equal to zero and the previous recurrence relation can be rewritten as $a_{-1}=1,\ a_0=-\frac12$ and $$\frac{1}{(2n+1)!}-\frac{1}{2(2n)!}+\sum_{k=1}^n\frac{a_{2k-1}}{(2n-2k+1)!}=0\quad(n\ge1).$$

Remark. The coefficients $B_n$ of the Taylor expansion $$\frac{z}{e^z-1}=\sum_{n\ge0}\frac{B_nz^n}{n!}$$ are known as Bernoulli numbers. Thus $a_n=\frac{B_{n+1}}{(n+1)!}$ for every $n\ge-1$.