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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$\begingroup$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 $tan(x)$. (Check the List of Maclaurin series of some common functions)
Proof: $$\frac{t}{e^t-1}=\frac{t}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1+\frac{(1-1)t-\frac{t^2}{2!}-\frac{t^3}{3!}-\frac{t^4}{4!}-...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$$\frac{t}{e^t-1}=1-\frac{t}{2}+\frac{+(\frac{1}{2}-\frac{1}{2!})t^2+(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{1}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{t}{2}+\frac{(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{t^4}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$$\frac{t}{e^t-1}=1-\frac{1}{2}t+\frac{\frac{1}{12}t^3+\frac{1}{24}t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{1}{2}t+\frac{1}{12}t^2+\frac{(\frac{1}{24}-\frac{1}{2.12})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$\endgroup$ 1 $\begingroup$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:
$$x\alpha(x) = e^{x} - 1.$$
$\endgroup$ $\begingroup$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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