# Is this formula correct, series and 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?

• You may want to check the sign of $x^2$ in the expansion of the 2nd equation. Also, on the same equation, where did the zero come from in the coefficient of $x^6$? – NoChance Nov 4 '19 at 0:15
• @NoChance: the third Bernoulli number is $0$ – James Warthington Nov 4 '19 at 0:21
• OK, May be you want to check the post below, look for the image with the formula #1541. The formula uses Even powers only - This may justify the inconsistent representation you found: math.stackexchange.com/questions/1885685/… – NoChance Nov 4 '19 at 0:26
• @NoChance: Do you know how can I expand this function into the series like Wolfram? Is there a way to expand this function and obtain the Bernoulli number before simplifying to get the equation in Wolfram? – James Warthington Nov 4 '19 at 0:31
• I could only do Taylor's expansion, but since the calculations are tedious, I am not sure what we will get. This link may be relevant: math.stackexchange.com/questions/3420676/… – NoChance Nov 4 '19 at 0:38

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*}