# Proving the Bernoulli number relation $(1+B)^n=B^n$

We know that we can generate the Bernoulli numbers using the relation $(1+B)^n=B^{[n]}$ where $B_n$ is $n$th Bernoulli number. But how we can prove this works? Thanks to all.

Edit 2: is there a website or book that can give me good information?

• If $B_n$ is a Bernoulli number then what is $B$? What is $B^{[n]}$? May 23, 2013 at 18:34
• @SohamChowdhury for example :$(1+B)^2=B^{[2]}$ so $1+2B^{[1]}+B^{[2]}=B^{[2]}$ so $1+2B_1+B_2=B_2$ so $B_1=-1/2$
– mnsh
May 23, 2013 at 18:46
• that way is only for Bernoulli number
– mnsh
May 23, 2013 at 18:47
• is there any website or book can give me good information ?
– mnsh
Jun 21, 2013 at 1:08
• As a possible "good information": you might look at/like go.helms-net.de/math/pascal/bernoulli_en.pdf , very elementary treatize which was just motivated by that expression with the Bernoulli-numbers. It was my first deeper encounter with number theory so please don't mind that it is much amateurish. Jun 21, 2013 at 5:06

The identity can be proven using generating functions. We have

$$1=\frac{t}{e^t-1}\frac{e^t-1}{t}=\left(\sum_{k=0}^\infty B_k\frac{t^k}{k!}\right)\left(\sum_{m=0}^\infty\frac{t^m}{(m+1)!}\right)=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)t^n.$$

Comparing coefficients of both sides yields for $$n\ge1$$:

$$\sum_{k=0}^n\frac{1}{k!(n-k+1)!}B_k=0\iff \sum_{k=0}^n\binom{n+1}{k}B_k=0\iff \sum_{k=0}^{n+1}\binom{n+1}{k}B_k=B_{n+1}.$$

Formally this is the relation $$(B+1)^{n+1}=B^{n+1}$$ expanded via binomial theorem then with the powers taken from superscript to subscript. The identity sometimes takes the recursive form

$$B_n=-\frac{1}{n+1}\sum_{k=0}^{n-1}\binom{n+1}{k}B_k.$$

This proof is present in these notes on Bernoulli numbers in the section on basic properties.

There are many resources available on GF techniques, notably generatingfunctionology.