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?

  • $\begingroup$ If $B_n$ is a Bernoulli number then what is $B$? What is $B^{[n]}$? $\endgroup$ May 23, 2013 at 18:34
  • $\begingroup$ @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 $B_1=-1/2$ $\endgroup$
    – mnsh
    May 23, 2013 at 18:46
  • $\begingroup$ that way is only for Bernoulli number $\endgroup$
    – mnsh
    May 23, 2013 at 18:47
  • $\begingroup$ is there any website or book can give me good information ? $\endgroup$
    – mnsh
    Jun 21, 2013 at 1:08
  • $\begingroup$ 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. $\endgroup$ Jun 21, 2013 at 5:06

1 Answer 1


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


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.


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