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 $\frac{-1}{2}$?
Best Answer
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 $z/(e^z-1)$.