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!
Best Answer
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: $$\frac{t}{e^t-1}=\frac{t}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1+\frac{(1-1)t-\frac{t^2}{2!}-\frac{t^3}{3!}-\frac{t^4}{4!}-...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$$\frac{t}{e^t-1}=1-\frac{t}{2}+\frac{+(\frac{1}{2}-\frac{1}{2!})t^2+(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{1}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{t}{2}+\frac{(\frac{1}{2.2!}-\frac{1}{3!})t^3+(\frac{1}{2.3!}-\frac{t^4}{4!})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$
$$\frac{t}{e^t-1}=1-\frac{1}{2}t+\frac{\frac{1}{12}t^3+\frac{1}{24}t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}=1-\frac{1}{2}t+\frac{1}{12}t^2+\frac{(\frac{1}{24}-\frac{1}{2.12})t^4+...}{t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...}$$