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: 

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

Proving two formulae involving two trigonometric series of Euler using Euler identities

I have recently read the book "Тригонометрические ряды от Эйлера до Лебега" (Trigonometric series from Euler to Lebesgue) and on page 12, there are 2 intriguing formlae derived by Euler in 1748. The book doesn't specify from which publication these two series come. Further up on that page, there is a mention of the "введения в анализ бесконечных" (Introduction to the analysis of the infinite), which was published also in 1748. Thus I suspect that these two formulae may come from there.

The above two formulae are as followed:

$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=1+r\cos(x)+r^2\cos(2x)...=1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)\tag{1}$$

$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)...=1+\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)\tag{2}$$

In the above book, the author states that Euler didn't provide the radius of convergence for these series, he wrote: "никаких указаний на области применимости этих и подобных им разложений у Эйлера нет" (No indications of range of applications are provided by Euler for these and similar expressions) (page 13)

The radius of convergence should be $|r|<1$ 

At $r=1$ and $r=-1$, Euler obtained:

$$-\dfrac{1}{2}=\cos(x)+\cos(2x)+\cos(3x)+...\tag{3}$$

$$\dfrac{1}{2}=\cos(x)-\cos(2x)+\cos(3x)+...\tag{4}$$

Both of these trigonometric series diverge $\forall x \in \mathbb{R}$

I wish to provide a derivation of these two formulae. This is derived from an answer of Student A Level, a member of Mathstackexchange, the site is here

We proceed by working with the RHS of the equation $(1)$ and $(2)$ and try to turn it into the LHS.

We denote the series in $(1)$ as $C$, and the series in $(2)$ as B

$$C=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\$$

$$S=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\$$

We will use the Euler identity $\color{green}{e^{ix}=\cos(x)+i\sin(x)}$ and $\color{green}{e^{inx}=\cos(nx)+i\sin(nx)}$

We multiply $S$ with imaginary number $i=\sqrt{-1}$, and obtain the new series:

$$iS=ri\sin(x)+r^2i\sin(2x)+r^3i\sin(3x)...$$

Summing these two series, we have

$$C+iS=1+r[\cos(x)+i\sin(x)]+r^2[(\cos(2x)+i\sin(2x)]+r^3[\cos(3x)+i\sin(3x)]...$$

Using Euler identity, we have:

$$C+iS=1+re^{ix}+(re^{ix})^2+(re^{ix})^{3}+(re^{ix})^4+...$$

This is, in fact, a geometric series, and the general formula is:

$$\dfrac{1}{1-kt}=1+kt+(kt)^2+(kt)^3+(kt)^4+...$$

Let $r=k$ and $t=e^{ix}$

We have $$\dfrac{1}{1-re^{ix}}$$

Let $r=k$ and $t=e^{ix}$

We have $$\dfrac{1}{1-re^{ix}}=1+re^{ix}+(re^{ix})^2+(re^{ix})^{3}+(re^{ix})^4+...$$

On the LHS: we mutiply the fraction with $1-re^{-ix}$, so we will have

$$\dfrac{1-re^{-ix}}{(1-re^{ix})(1-re^{-ix})}=\dfrac{1-r(\cos(x)-ir\sin(x))}{1-re^{-ix}-re^{ix}+r^2}$$

This is the tricky part of the proof, we wish to turn

$$1-re^{-ix}-re^{ix}+r^2=1-r(e^{-ix}+e^{ix})+r^2=1-2r\cos(x)+r^2$$

We can work backward and realize that $\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$, so we need to multiply $re^{-ix}-re^{ix}$ with $2r\left(\dfrac{e^{-ix}+e^{ix}}{2}\right)$. So we will have:

$$1-re^{-ix}-re^{ix}+r^2=1-2r\left(\dfrac{e^{-ix}+e^{ix}}{2}\right)+r^2=1-2r\cos(x)+r^2$$

Thus we have:

$$\dfrac{1-r(\cos(x)-i\sin(x)}{1-2r\cos(x)+r^2}=\dfrac{1-r\cos(x)+ir\sin(x)}{1-2r\cos(x)+r^2}=\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}+\dfrac{ir\sin(x)}{1-2r\cos(x)+r^2}$$

Equating the imaginary and the real part, we thus have:

$$\boxed{\color{green}{A=\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)}}$$

$$\boxed{\color{green}{B=\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)}}$$