Deriving the power series of the generating function of the Bernoulli numbers.

In this thread, we are going to derive the series of $\dfrac{x}{e^x-1}$. This series is especially important because it is the generating function of the Bernoulli numbers. I have tried many times in vain to derive the series, but with the precious help of members of Mathstack Exchange, I could finally derive it.

The series is especially difficult because the $n$ order derivative is very complex as we progress. For example:

$$f(x)=\dfrac{x}{e^x-1} \implies f'(x)=\dfrac{e^x(x-1)+1}{(e^x-1)^2}\implies f'(0)=\dfrac{1(0-1)+1}{1-1}=\dfrac{1}{0}$$

We can of course rely on the L'Hopital rule to calculate the limit of $x$ approaches $0$ in all $n$ order derivatives but it is a hugely complicated calculation.

Another way must be found to derive the series for this function. The method is to rely on the equality: $$\left(\dfrac{x}{e^x-1}\right)\left(\dfrac{e^x-1}{x}\right)=1\tag{1}$$

We can easily expand $\dfrac{e^x-1}{x}$ into its power series by using the series $e^x=\sum_{n=0}^{+\infty}\dfrac{x^n}{n!}$

$$\dfrac{e^x-1}{x}=\dfrac{\color{red}{1}+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...\color{red}{-1}}{x}=1+\dfrac{x}{2!}+\dfrac{x^2}{3!}+\dfrac{x^3}{4!}+...=\sum_{n=1}^{+\infty} \dfrac{x^{n-1}}{n!}=\sum_{n=0}^{+\infty} \dfrac{x^{n}}{(n+1)!}\tag{2}$$

Now, denote the series of $\dfrac{x}{e^x-1}=\sum_{n=0}^{+\infty}b_nx^n$ and the series of $\dfrac{e^x-1}{x}=\sum_{n=0}^{+\infty} a_nx^n$.

The coefficient of the $\sum_{n=0}^{+\infty} a_nx^n$ is $a_n=\dfrac{1}{(n+1)!}\tag{3}$

We can rewrite the equation in $(1)$ as followed:

$$\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...\right)\sum_{n=0}^{+\infty}b_nx^n=1+0x+0x^2+0x^3+...\tag{4}$$

$$=\left(\sum_{n=0}^{+\infty}b_nx^n\right)\left(\sum_{n=0}^{+\infty}a_nx^n\right)=1\tag{5}$$

We can compare the product of the coefficients $b_n$ and $a_n$ on LHS to the coefficients of the "series" on the RHS in $(4)$ and form this system of equations:

$[x^0]a_0b_0=1$

$[x^1]a_0b_1+a_1b_0=0$

$[x^2]a_0b_2+a_1b_1+a_2b_0=0$

$[x^3]a_0b_3+a_1b_2+a_2b_1+a_3b_0=0$

$[x^4]a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0=0$

$[x^5]a_0b_5+a_1b_4+a_2b_3+a_3b_2+a_4b_1+a_5b_0=0$

$[x^6]a_0b_6+a_1b_5+a_2b_4+a_3b_3+a_4b_2+a_5b_1+a_6b_0=0$

Refer to this thread to review what is a Cauchy triangle when multiplying 2 power series:

From the system of equations above, we can easily deduce the value for $b_n$:

$a_0=1$

$b_0=1/a_0=\dfrac{1}{1}=1$

$b_1=-a_1b_0/a_0=-\left(\dfrac{1}{2}\cdot1\right)=-\dfrac{1}{2}$

$b_2=-(a_2b_0+a_1b_1)/a_0=-\left(\dfrac{1}{3!}\cdot1+\dfrac{1}{2!}\cdot\left(-\dfrac{1}{2}\right)\right)=\dfrac{1}{12}$

$b_3=-(a_3b_0+a_2b_1+a_1b_2)/a_0=-\left(\dfrac{1}{4!}\cdot1+\dfrac{1}{3!}\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}\cdot\left(\dfrac{1}{12}\right)\right)=0$

$b_4=-(a_4b_0+a_3b_1+a_2b_2+a_1b_3)/a_0=-\left(\dfrac{1}{5!}\cdot1+\dfrac{1}{4!}\left(-\dfrac{1}{2}\right)+\dfrac{1}{3!}\\\cdot\dfrac{1}{12}+\left(\dfrac{1}{2}\cdot0\right)\right)=-\dfrac{1}{720}$

If we continue on, we will find that $b_n$ when $n$ is odd are all $0$, except that $b_1=-\dfrac{1}{2}$

We only need to compute for $b_n$ when n is even, and these are even values of $b_n$

$b_0=1$

$b_2=-\dfrac{1}{2}$

$b_4=-\dfrac{1}{720}$

$b_6=\dfrac{1}{30240}$

$b_8=-\dfrac{1}{1209600}$

$b_{10}=\dfrac{1}{47900160}$

So the coefficients for the series of $\dfrac{x}{e^x-1}$ will be:

$$\dfrac{x}{e^x-1}=1-\dfrac{1}{2}x+\dfrac{1}{12}x^2-\dfrac{1}{720}x^4+\dfrac{1}{30240}x^6-\dfrac{1}{1209600}x^8+\dfrac{1}{47900160}x^{10}+...$$

To make the Bernoulli numbers appear, we divide the denominator by $n!$. For example, for the fourth term , we divide $720$ by $4!$ and we obtain $\dfrac{1}{30}$, which the fourth Bernoulli number. We can rewrite:

$$=\color{green}{1}-\dfrac{\color{green}{1}}{\color{green}{2}\cdot1!}x+\dfrac{\color{green}{1}}{\color{green}{6}\cdot2!}x^2-\dfrac{\color{green}{1}}{\color{green}{30}\cdot4!}x^4+\dfrac{\color{green}{1}}{\color{green}{42}\cdot6!}x^6-\dfrac{\color{green}{1}}{\color{green}{30}\cdot8!}x^8+\dfrac{\color{green}{5}}{\color{green}{66}\cdot10!}x^{10}+...$$

The numbers colored in green are Bernoulli numbers.

Thus the general formula of the series is: 

$$\dfrac{x}{e^x-1}=\sum_{n=0}^{+\infty}\dfrac{B_nx^n}{n!}$$