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?
