how the following formula or any other explicit formula for computing Bernoulli numbers can be derived?

Question

how the following formula $$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$ can be derived? I know how to use the formula but still have not seen any proof for the given formula.

Answer 1

The recurrence relation \begin{align*} B_n&=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}\qquad n\geq 0\\ \end{align*}

specifies a variation of the Bernoulli numbers with $B_1=+\frac{1}{2}$ instead of the more common value $B_1=-\frac{1}{2}$.

We therefore start with the generating function \begin{align*} \sum_{k=0}^\infty \frac{B_k}{k!}x^k=\frac{x}{e^x-1}+x=\frac{xe^x}{e^x-1}\tag{1} \end{align*}

Note: The recurrence relation also gives $B_0=1$ since we use the convention that an empty sum is equal to zero. This is the case for $n=0$ when the upper limit of the sum is less than the lower limit.

Multiplication of (1) by $e^x-1$ gives

\begin{align*} xe^x&=\left(\sum_{k=0}^\infty\frac{B_k}{k!}x^k\right)\left(e^x-1\right)\tag{2}\\ &=\left(\sum_{k=0}^\infty\frac{B_k}{k!}x^k\right)\left(\sum_{l=1}^\infty\frac{x^l}{l!}\right)\tag{3}\\ &=\sum_{n=1}^\infty \left(\sum_{{k+l=n}\atop{k\geq 0,l\geq 1}}\frac{B_k}{k!}\frac{1}{l!}\right)x^n\tag{4}\\ &=\sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}\frac{B_k}{k!}\frac{1}{(n-k)!}\right)x^n\tag{5}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)x^{n+1}\tag{6}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\right)\frac{x^{n+1}}{n!}\tag{7}\\ \end{align*}

Comment:

• In (3) we expand the exponential function $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$.

• In (4) we do the Cauchy multiplication of series.

• In (5) we eliminate $l$ by using $l=n-k$.

• In (6) we shift the index $n$ to start with $n=0$.

• In (7) we expand numerator and denominator by $n!$ and introduce the binomial coefficient $\binom{n}{k}$.

From (7) and the expansion of the left-hand side of (2) we obtain after division by $x$

\begin{align*} \sum_{n=0}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\right)\frac{x^{n}}{n!}\tag{8} \end{align*}

Comparison of the coefficient of $x^n$ gives after multiplication with $n!$ for $n\geq 0$:

\begin{align*} \color{blue}{1}&=\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\\ &\,\,\color{blue}{=B_n+\sum_{k=0}^{n-1}\binom{n}{k}\frac{B_k}{n-k+1}} \end{align*}

and the claim follows.

Answer 2

It is immediate from the definition $$\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{B_k}{k!} x^k$$ which means $$1=\frac{e^x-1}{x} \sum_{k=0}^\infty \frac{B_k}{k!} x^k = ( \sum_{m=0}^\infty \frac{1}{(m+1)!} x^m)( \sum_{k=0}^\infty \frac{B_k}{k!} x^k) = \sum_{n=0}^\infty (\sum_{k=0}^n \frac{B_k}{k!}\frac1{ (n-k+1)!}) x^n$$ ie. $$B_0=1, \qquad for \ n > 1\ \qquad \sum_{k=0}^n \frac{B_k}{k! (n-k+1)!} = 0$$

Answer 3

The following formula has been proved in The American Mathematical Monthly, but here the proof is deeper and doesn't use any induction:

consider the function $$f\left(x\right)=\frac{1}{e^{x}+1}=\frac{e^{x}+1-2}{e^{2x}-1}=\frac{1}{e^{x}-1}-\frac{2}{e^{2x}-1}=\frac{1}{x}\cdot\frac{x}{e^{x}+1}-\frac{1}{x}\cdot\frac{2x}{e^{2x}-1}$$

Using the famous definition of Bernoulli numbers we have:

$$f\left(x\right)=\frac{1}{x}\sum_{n=0}^{∞}B_{n}\ \frac{x^{n}}{n!}-\frac{1}{x}\sum_{n=0}^{∞}B_{n}\ \frac{2^{n}x^{n}}{n!}=\sum_{n=0}^{∞}B_{n}\ \frac{x^{n-1}}{\left(n\right)!}\left(1-2^{n}\right)$$

let $n-1↦n$, implies:$$f\left(x\right)=\sum_{n=0}^{∞}B_{n+1}\ \frac{x^{n}}{\left(n+1\right)!}\left(1-2^{n+1}\right)$$

Computing $$f^{\left(n\right)}\left(0\right)$$ gives the following relation: (I) $$f^{\left(n\right)}\left(0\right)=\frac{B_{n+1}\left(1-2^{n+1}\right)}{n+1}$$

On the other hand from the answer of user90369 we can compute the $n^{th}$ derivative of $$ \frac{1}{e^{x}+1}$$ around $x=0$ which is as follows:

(II) $$\frac{d^{n}}{dx^{n}}\frac{1}{e^{x}+1}=\sum_{k=0}^{n}\left(-1\right)^{n}\sum_{j=0}^{k}\left(-1\right)^{j}{{k}\choose{j}}\left(j+1\right)^{n}\cdot\frac{1}{2^{k+1}}$$

comparing the $n^{th}$ derivative of the function from I and II we arrive at the explicit formula for the $n+1^{th}$ Bernoulli number:

$$B_{n+1}=\frac{\left(-1\right)^{\left(n+1\right)}\left(n+1\right)}{2^{\left(n+1\right)}-1}\sum_{k=1}^{n+1}\frac{1}{2^{\left(k\right)}}\sum_{j=0}^{k-1}\left(-1\right)^{j}\left(j+1\right)^{n}{{k-1}\choose{j}}$$ which is valid for $n\ge0$.