How can I calculate closed form of a sum? As we know the closed form of
$$
{1 \over k} {n \choose k - 1}
\sum_{j = 0}^{n + 1 - k}\left(-1\right)^{\, j}
{n + 1 - k \choose j}B_{j} =
\bbox[10px,border:2px solid #00A000]{{n! \over k!}\, 
\left[\, z^{n + 1 - k}\,\,\,\right]
{z \exp\left(z\right) \over 1 - \exp\left(-z\right)}}
$$
where $B_{n}$ the Bernoulli sequence defined by the exponential generating function
$$
{x \over \mathrm{e}^{x} - 1} = \sum_{n = 0}^{\infty}{B_{n} \over n!}\,x^{n}
$$
The problem is here !. How can I get to this closed form ?.
 A: 
We obtain
\begin{align*}
\frac{ze^z}{1-e^{-z}}&=\frac{-z}{e^{-z}-1}\cdot e^z\\
&=\left(\sum_{j=0}^\infty\frac{B_j}{j!}(-z)^j\right)\left(\sum_{k=0}^\infty\frac{z^k}{k!}\right)\tag{1}\\
&=\sum_{n=0}^\infty \left(\sum_{{j+k=n}\atop{j,k\geq 0}}\frac{B_j}{j!}(-1)^j\frac{1}{k!}\right)z^n\\
&=\sum_{n=0}^\infty \left(\sum_{j=0}^n\binom{n}{j}B_j(-1)^j\right)\frac{z^n}{n!}\tag{2}\\
\end{align*}

Comment:


*

*In (1) we expand the exponential generating function $\frac{z}{e^z-1}$  of the Bernoulli numbers and substitute $z\rightarrow -z$.


We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series.

We obtain from (2)
  \begin{align*}
\frac{n!}{k!}[z^{n+1-k}]\frac{ze^z}{1-e^{-z}}
&=\frac{n!}{k!}\sum_{j=0}^{n+k-1}\binom{n+1-k}{j}B_j(-1)^j\frac{1}{(n-k+1)!}\tag{3}\\
&=\frac{n!}{k!(n-(k-1))!}\sum_{j=0}^{n+1-k}\binom{n+1-k}{j}B_j(-1)^j\\
&=\frac{1}{k}\binom{n}{k-1}\sum_{j=0}^{n+1-k}\binom{n+1-k}{j}B_j(-1)^j\\
\end{align*}
  and the claim follows.

Comment:


*

*In (3) we select the coefficient of $z^{n+1-k}$ from (2).


Note:  Usually we don't call the RHS of the summation formula a closed form but instead the coefficient of its generating function.
