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.


  • 3
    $\begingroup$ 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. $\endgroup$
    – anon
    Jul 31, 2013 at 4:45
  • $\begingroup$ @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 $\frac{t}{e^t-1}=\sum_{n=1}^{\infty} B_n \frac{t^n}{n!}$ $\endgroup$
    – PPP
    Jul 31, 2013 at 5:09
  • $\begingroup$ 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. $\endgroup$
    – anon
    Jul 31, 2013 at 5:20
  • $\begingroup$ @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 $\endgroup$
    – PPP
    Jul 31, 2013 at 5:25
  • 1
    $\begingroup$ @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 $\endgroup$
    – PPP
    Aug 1, 2013 at 4:49

4 Answers 4


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

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


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

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



  • 1
    $\begingroup$ So, Bernoulli's work was only empirical? $\endgroup$
    – 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 $$ C_0=1,\quad C_{n-1}=0\quad \text{for} \quad n>1 $$


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