Proving that Bernoulli Number $B_0 =1$

Question

This question was given by our instructor today to do by ourselves and I am having trouble proving it. So, I am asking here.

Definition of Bernoulli Polynomials : For any complex number x we define the function $B_n(x) $ by the equation $\frac{z e^{xz} } { e^z -1} = \sum_{n=0}^{\infty} \frac{B_n(x) } {n! } z^n$ , where $|z|<2π $.

$B_n(0)$ are called Bernoulli numbers and are denoted by $B_n$.

I have to prove that $B_0$ is $1$.

I tried by putting $x=0$ and then $n=0$ , so I get $ \frac {z}{(e^{z}-1) } = B_n $ but not what value should I take on LHS of $z$; every value $|z|<2π$ will be fine as it's true for definition.

If I take $z=1$ then it will not work as I get LHS as $1/(e-1)$ .

I have been trying this for half an hour but couldn't solve it and tried two hours earlier also.

Please help!!

Answer 1

I think you should get $\dfrac z{e^z-1}=\sum\limits_{n=0}^{\infty}\dfrac{B_n}{n!}z^n$ rather than what you wrote (namely, $=B_n$).

Then $z=\left(e^z-1\right)\left(\sum\limits_{n=0}^{\infty}\dfrac{B_n}{n!}z^n\right)=\left(z+\dfrac{z^2}{2!}+\cdots\right)\left(B_0+B_1z+\dfrac{B_2}{2!}z^2+\cdots\right).$

Now can you see that $B_0=1$?

Answer 2

The Bernoulli numbers $B_n$ are generated by \begin{equation}\label{BN-dfn}\tag{1} \frac{z}{e^z-1}=\sum_{n=0}^\infty B_n\frac{z^n}{n!}, \quad \vert z\vert<2\pi. \end{equation} Because the function $\frac{x}{e^x-1}-1+\frac{x}2$ is even in $x\in\mathbb{R}$, all of the Bernoulli numbers $B_{2k+1}$ for $k\in\mathbb{N}$ equal $0$.

The Maclaurin series in \eqref{BN-dfn} implies \begin{equation} B_n=\lim_{z\to0}\biggl(\frac{z}{e^z-1}\biggr)^{(n)}, \quad n\ge0. \end{equation} Hence, it is trivial that \begin{equation} B_0=\lim_{z\to0}\biggl(\frac{z}{e^z-1}\biggr)^{(0)} =\lim_{z\to0}\frac{z}{e^z-1} =1. \end{equation} For more information on computation of the Bernoulli numbers $B_n$, please click at the site https://math.stackexchange.com/a/4254493/945479.