# 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!

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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 '13 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 $\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$ – Lucas Zanella Jul 31 '13 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 '13 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 – Lucas Zanella Jul 31 '13 at 5:25
"I'm more interested in a proof that doesn't assume the formula before the proof." You just told me you thought of this formula as the definition of the Bernoulli numbers. Now you're saying you don't want the formula to be assumed? That's contradictory. – anon Jul 31 '13 at 7:07

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

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

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

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So, Bernoulli's work was only empirical? – Lucas Zanella Feb 17 '14 at 4:32