An equation by the definition of Bernoulli number

Question

I am working on Bernoulli number. I learnt the definition of Bernoulli number on the book by a Japanese mathematician. The name of the book is Number Theory 1: Fermat's dream. The book defines the Bernoulli number by the formula $\dfrac{x}{e^x-1}=\sum\limits_{n=0}^\infty \dfrac{B_n}{n!}x^n$. And it defines a linear operator $D:\mathbb{C}[x]\rightarrow\mathbb{C}[x]:f(x)\rightarrow \dfrac{d}{dx}f(x)$ and $e^D=\sum\limits_{n=0}^\infty\dfrac{D^n}{n!}$. And it says that $D=(e^D-1)\sum\limits_{n=0}^\infty\dfrac{B_n}{n!}D^n$. My confusion is that why the last equation is right?

Answer 1

By definition and using Cauchy product, $$ x = (e^x - 1)\sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}x^n } = \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{n!}}} \sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}x^n } = \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^{n-1} {\binom{n}{k}B_k } } \right)\frac{{x^n }}{{n!}}} . $$ Thus, comparing the coefficients of like powers of $x$, $$\tag{1} \sum\limits_{k = 0}^{n-1} {\binom{n}{k}B_k } = 0 $$ if $n\neq 1$ and when $n=1$, $B_0=1$. Hence, $$ (e^D - 1)\sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}D^n } = \sum\limits_{n = 1}^\infty {\frac{{D^n }}{{n!}}} \sum\limits_{n = 0}^\infty {\frac{{B_n }}{{n!}}D^n } = \sum\limits_{n = 0}^\infty {\left( {\sum\limits_{k = 0}^{n-1} {\binom{n}{k}B_k } } \right)\frac{{D^n }}{{n!}}} =D $$ where we used $(1)$ in the last step.