Question

How is it going?
Please, explain me

Answer 1

\(z= \dfrac{z}{e^z -1}(e^z-1)\)
but
\(e^z-1) =\sum_{n=1}^\infty \dfrac{x^k}{k!}\)

Answer 2

and \(\dfrac{z}{e^z-1}=\sum_{m=0}^\infty \dfrac{B_m}{m!}z^n\)

Answer 3

Hence,
\(z= \sum_{n=0}^\infty (\sum_{k=1}^\infty \dfrac{B_{n-k}}{k!(n-k)!})z^n)\)

Answer 4

why?

Answer 5

Then: \(\sum_{k=1}^n \dfrac{B_{n-k}}{k!(n-k)!} =\begin{cases}0~~if~~n\neq 1\\1~~if~~n=1\end{cases}\)
Again, why?

Answer 6

Sorry, typo at \(e^z-1=\sum_{k=1}^\infty \dfrac{z^k}{k!}\)