# Bernoulli Numbers

I've read that Bernoulli Numbers are defined by the series $$\frac{z}{e^z-1}\equiv \sum\limits_{n=0}^{\infty}B_n\frac{z^n}{n!},$$
So if I start with $0$ I get $$B_0\frac{1}{1}=B_0{1}.$$ My question is, why is there a $B_0$ in the term...is it of any significance? Or just a "marker" or something to indicate that this is the $B_0$ term?

If I find the second term I get $$B_1\frac{z}{1}=B_1z$$ What's the $z$? I've read it must be $\left|z\right|<2\pi$, but how does one get $\frac{-1}{2}$?

• In other words, couldn't this second term be virtually anything in $-2\pi<z<2\pi$? – bjd2385 Oct 2 '14 at 12:29
• I think that $B_0=1$ and $B_1=-\frac{1}{2}$ are defined by convention for the first Bernoulli numbers. $B_0=1$ and $B_1=+\frac{1}{2}$ define the second Bernoulli numbers. – Claude Leibovici Oct 2 '14 at 12:40

First of all, using the Taylor series for $e^z$ we have $$\frac{e^z-1}{z} = 1 + \frac{z}{2} + \frac{z^2}{6} + \frac{z^3}{24} + \cdots.$$ Multiplying this by the power series for $z/(e^z-1)$ and comparing coefficients (the product should be $1$) we get \begin{align*} 1 &= B_0 \\ 0 &= B_1 + 1/2 \\ 0 &= (2B_2) + (1/2) B_1 + (1/6) B_0 \\ 0 &= (6B_3) + (1/2) (2B_2) + (1/6) B_1 + (1/24) B_0 \end{align*} and so on. Therefore $B_0 = 1$, $B_1 = -1/2$, $B_2 = 1/6$, $B_3 = -1/30$, and so on.
If you look at the function $$f(z) = \frac{z}{e^z-1} + \frac{z}{2}$$ then you find out that \begin{align*} f(-z) = \frac{-z}{e^{-z}-1} - \frac{z}{2} = \frac{ze^z}{e^z-1} - \frac{z}{2} = \frac{z}{e^z-1} + z - \frac{z}{2} = f(z). \end{align*} Therefore $f(z)$ is even and all the odd coefficients in its power series vanish. This shows that apart from $B_1 = -1/2$, all other odd-indexed Bernoulli numbers vanish. Why do we need them, then? They're just the sequence whose exponential generating series is $z/(e^z-1)$.
The statement defining the Bernoulli numbers says that the power series expansion of $\frac{z}{e^z-1}$ is equal to the power series on the right; that is, the coefficient of $x^i$ in the power series expansion is $B_i$. The first few terms of the expansion are $$\frac{z}{e^z-1} = 1-\frac{z}{2}+\frac{z^2}{12}-\frac{z^4}{720}+\frac{z^6}{30240}+\cdots.$$ Thus \begin{align*} B_0&=1 \\ B_1&=-\frac{1}{2}\\ B_2&=2!\cdot\frac{1}{12}=\frac{1}{6}\\ B_3&=0\\ B_4 &= -4!\cdot\frac{1}{720} = -\frac{1}{30} \end{align*} and so forth.