Is this formula correct, series and Bernoulli numbers

Question

According to this page: https://proofwiki.org/wiki/Definition:Bernoulli_Numbers

The series is : $\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{B_1x^2}{2!}+\dfrac{B_2x^4}{4!}+\dfrac{B_3x^6}{6!}...$ where $B_1=-\dfrac{1}{2}, B_2=\dfrac{1}{6}, B_3=0, B_4=-\dfrac{1}{30}$

How can I expand obtain the Bernoulli coefficient as indicated in the website?

Using the series above, I have:

$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{(-\frac{1}{2})x^2}{2!}+\dfrac{\frac{1}{6}x^4}{4!}+\dfrac{(0)x^6}{6!}...$

$=\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{4}+\dfrac{x^4}{144}+0...$

This looks wrong since the series expansion according to Wolfram is:

$\dfrac{x}{e^x-1}=1-\dfrac{x}{2}+\dfrac{x^2}{12}-\dfrac{x^4}{720}...$

But then the series of Wolfram doesn't show the Bernoulli number. I am confused, can you help me here?

Answer 1

Note, the generating function for the Bernoulli numbers is defined as \begin{align*} \frac{x}{e^x-1}=\sum_{n=0}^\infty\frac{B_nx^n}{n!} \end{align*} with $\frac{B_\color{blue}{n}}{n!}$ the coefficient of $x^\color{blue}{n}$.

Since we have $B_0=1, B_1=-\frac{1}{2}, B_2=\frac{1}{6}, B_3=0, B_4=-\frac{1}{30},\ldots$ we obtain \begin{align*} \color{blue}{\frac{x}{e^x-1}}&=\sum_{n=0}^\infty\frac{B_nx^n}{n!}\\ &=B_0+\frac{B_1x}{1!}+\frac{B_2x^2}{2!}+\frac{B_3x^3}{3!}+\frac{B_4x^4}{4!}+\cdots\\ &=1+\frac{\left(-\frac{1}{2}\right)x}{1}+\frac{\left(\frac{1}{6}\right)x^2}{2}+\frac{0x^3}{6}+\frac{\left(-\frac{1}{30}\right)x^4}{24}+\cdots\\ &\,\,\color{blue}{=1-\frac{x}{2}+\frac{x^2}{12}-\frac{x^4}{720}+\cdots} \end{align*}