# Bernoulli numbers generating function

Consider the following generating formula:

$$\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$$

There is some intuitive explanation about it?

I want to know because I need to proof to myself that the sum of the combination of the Bernoulli Numbers is $0$, like this: $$\sum_{u=1}^\infty {{n+1}\choose u} B_u = 0$$ I've already understood the entire proof, but it assumes that $\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$ so I want to proof (or see how it was found) this last part.

Thanks!

• Generally that's a definition of the Bernoulli numbers. If that's not the definition for you, and neither is the recurrence relation, then you'll have to specify what definition you are operating on.
– anon
Jul 31, 2013 at 4:45
• @anon my definition of bernoulli numbers is that they are coefficients that create formulas for sum of powers. Assume that I'm Bernoulli and you need to teach me that $\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$
– PPP
Jul 31, 2013 at 5:09
• One can prove Faulhaber's formula in terms of the coefficients of $\frac{t}{e^t-1}$ as an exponential generating function. Since Faulhaber's formula uniquely define the Bernoulli numbers, this proves these coefficients and the $B_n$ numbers are one and the same.
– anon
Jul 31, 2013 at 5:20
• @anon but how did somebody find this equation? What passed in his head to show up this equation? What's the intuition behind this? I'm more intersted in a proof that doesn't assumes the formula before the proof
– PPP
Jul 31, 2013 at 5:25
• @anon yes, that's my definition. But I didn't say that the formula $\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$ is the definition of Bernoulli numbers. I'm talking about the formulas for sum of powers
– PPP
Aug 1, 2013 at 4:49

Let's assume that $g(x)$ is given and we try to find out $f(n)$

$$f(n)=\sum_{i=1}^n g(i)$$

$$f(n+1)=\sum_{i=1}^{n+1}g(i)$$

$$f(n+1)-f(n)=g(n+1) \tag 1$$

We know Taylor expansion

$$f(x+h)=f(x)+hf'(x)+\frac{h^2 f''(x)}{2!}+\frac{h^3f'''(x)}{3!}+....$$

Thus

$$f(n+1)=f(n)+f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+....$$

If we put $f(n+1)$ taylor expansion in Equation $1$

$$f(n+1)-f(n)=g(n+1)$$ $$f(n)+f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+....-f(n)=g(n+1)$$

$$f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+...=g(n+1) \tag 2$$

$$f(n)+\frac{f'(n)}{2!}+\frac{f''(n)}{3!}+\frac{f'''(n)}{4!}+...=\int g(n+1) dn$$

We need $f(n)$ if so we need to cancel $f'(n)$ . So we need to

$$-\frac{1}{2} ( f'(n)+\frac{f''(n)}{2!}+\frac{f'''(n)}{3!}+...)=-\frac{1}{2}g(n+1)$$

$$f(n)+ (-\frac{1}{2.2} +\frac{1}{3!})f''(n)+(-\frac{1}{2.3!} +\frac{1}{4!})f'''(n)+...=\int g(n+1) dn-\frac{1}{2}g(n+1)$$

$$f''(n)+\frac{f'''(n)}{2!}+\frac{f^{4}(n)}{3!}+...=\frac{d(g(n+1))}{dn}$$

If you continue in that way to cancel $f^{r}(n)$ terms step by step, you will get

$$f(n)=\int g(n+1) dn-\frac{1}{2}g(n+1)+\frac{1}{12}\frac{d(g(n+1))}{dn}+a_4\frac{d^2(g(n+1))}{dn^2}+a_5\frac{d^3(g(n+1))}{dn^3}+...$$

This is Euler-Maclaurin formula. (Please see also the Applications of the Bernoulli numbers). I just wanted to show Bernoulli numbers seen in one of the very important formulas in mathematics .

Where $$a_n= \frac{B_n}{n!}$$.

Because If you try to find out the coefficients of $\frac{t}{e^t-1}$ by polynomial division. You can get exactly same coefficients that seen in Euler-Maclaurin formula.

The Bernoulli numbers appear in Jacob Bernoulli's most original work "Ars Conjectandi" published in Basel in 1713 in a discussion of the exponential series.

You can also see that The Bernoulli numbers appears in the power series of $tan(x)$. https://en.wikipedia.org/wiki/Taylor_series (Check the List of Maclaurin series of some common functions)

Proof: $$\frac{t}{e^t-1}=\frac{t}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1+\frac{(1-1)t-\frac{t^2}{2!}-\frac{t^3}{3!}-\frac{t^4}{4!}-...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$

$$\frac{t}{e^t-1}=1-\frac{t}{2}+\frac{+(\frac{1}{2}-\frac{1}{2!})t^2+(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{1}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{t}{2}+\frac{(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{t^4}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$

$$\frac{t}{e^t-1}=1-\frac{1}{2}t+\frac{\frac{1}{12}t^3+\frac{1}{24}t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{1}{2}t+\frac{1}{12}t^2+\frac{(\frac{1}{24}-\frac{1}{2.12})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$

• So, Bernoulli's work was only empirical?
– PPP
Feb 17, 2014 at 4:32

How about a solution using exponential generating functions? Begin with the conditions $$B_0 = 1$$ and $$\sum_{k = 0}^{n} \binom{n+1}{k}B_k = 0$$. Rearrange the latter identity to get $$B_k = \frac{-1}{n+1}\sum_{k = 0}^{n-1}\binom{n+1}{k}B_k$$. Apply the coefficients of the exponential generating function to get $$F(x) = \sum_{n = 0}^{\infty}\frac{B_nx^{n}}{n!} = B_0 + \sum_{n = 1}^{\infty}\sum_{k = 0}^{n-1}\frac{B_kx^{k}}{k!}\cdot\frac{x^{n-k}}{(n-k+1)!}.$$ By associativity, we can change the order of the summation to get $$F(x) = 1 + \sum_{k = 0}^{\infty}\frac{B_kx^{k}}{k!}\sum_{n = 1}^{\infty} \frac{x^{n}}{(n+1)!}$$
(the easiest way to see this is to make a table with entries associated with each $$n$$ and $$k$$, then grouping the terms based on $$\frac{B_kx^{k}}{k!}$$). Notice that the stuff in the second summation has closed form $$\frac{e^{x}-1}{x} - 1$$ so $$F(x) = 1 + \left(1 - \frac{e^{x}-1}{x}\right)F(x).$$ Rearranging the identity in-terms of $$F(x)$$ gives you the desired closed form for the generating function of $$B_n$$.

One is interested in finding a closed form expression for the following power series:

$$\beta(x) = \sum^{\infty}_{k=0}\frac{B_{k}}{k!}x^{k}.$$

A natural way to arrive at such an expression is from the well-known recursive expression for the Bernoulli numbers:

$$\sum^{n}_{k=0}\binom{n+1}{k}B_{k} = \delta_{n,0},$$

for n = 0, 1, 2, 3, ... and $$\delta$$ is the Kronecker delta.

This recursive expression is very interesting because it suggests that $$\beta(x)$$ should be multiplied by a well-designed power series to yield a very simple one at the end.

Consider the following power series:

$$\alpha(x) = \sum^{\infty}_{k=0}a_{k}x^{k}.$$

Define $$\gamma(x)$$ to be the product of $$\alpha(x)$$ and $$\beta(x)$$ as follows:

$$\gamma(x) = \sum^{\infty}_{k = 0}g_{k}x^{k} = \alpha(x)\beta(x)$$.

From the formula for the coefficients of the product of two power series one gets:

$$g_{n} = \sum^{n}_{k = 0}a_{n-k}\frac{B_{k}}{k!}.$$

From the formula above and the recursive expression it is natural to define $$a_{n-k} = \frac{1}{(n + 1 - k)!}$$ so that $$a_{n} = \frac{1}{(n+1)!}$$ for $$n = 0, 1, 2, 3...$$ It is now easy to see that $$g_{n} = 0$$ for $$n= 1, 2, 3 ...$$ and $$g_{0} = 1$$.

From the considerations above one arrives at the following expression for $$\alpha(x)$$:

$$\alpha(x) = \sum^{\infty}_{k=0}\frac{1}{(k+1)!}x^{k},$$

hence, it is easy to see that:

$$x\alpha(x) = e^{x} - 1.$$

This is one of the problems in VLA, see problem 515.

Let us start with the generating function $$z=(e^z-1)\sum_{n=0}^\infty \frac{B_{n}}{n!}z^n,$$ where the expansion of $$(e^z-1)$$ gives $$(e^z-1)=\sum_{k=0}^\infty \frac{z^{k+1}}{(k+1)!}.$$ Then applying Cauchy product, we arrive at $$z=\sum_{n=0}^\infty \sum_{k=0}^n \frac{B_{k}}{k!}\frac{z^{n+1}}{(n-k+1)!}$$ or $$z=C_0 z+\sum_{n=2}^\infty C_{n-1} \frac{z^{n}}{n!}$$ where $$C_{n-1}=\sum_{k=0}^{n-1} \frac{B_{k}n!}{k!(n-k)!}.$$ Comparing the left side with right one, we get $$C_0=1,\quad C_{n-1}=0\quad \text{for} \quad n>1$$