An identity involving Bernoulli and Stirling numbers I was playing with some combinatorial sums and made an observation that I didn't know how to prove:
$$\forall n\in\mathbb N,\hspace{10px}\sum_{k=1}^n\frac{B_k\ S_1(n-1,\,k-1)}k=-\sum_{k=1}^n\frac{S_1(n,\,k)}{(k+1)\ n},$$
where $B_k$ are Bernoulli numbers and $S_1(n,\,k)$ are signed Stirling numbers of the first kind.
Could you please suggest any ideas how to prove it?
 A: Suppose we seek to show that
$$\sum_{k=1}^n \frac{B_k}{k} (-1)^{n-k} 
{n-1\brack k-1}
= -\frac{1}{n} 
\sum_{k=1}^n \frac{1}{k+1} (-1)^{n-k} {n\brack k}.$$
Recall the classic generating function  of the Stirling numbers of the
first kind which yields
$${n\brack k} = n! [z^n][u^k] \exp\left(u\log\frac{1}{1-z}\right).$$
and also
$${n\brack k} = n! [z^n] \frac{1}{k!}\left(\log\frac{1}{1-z}\right)^k$$
which controls the  range so that for $n\lt k$ we  get zero, and hence
we may let $k$ go to infinity in the two sums.

We get for the LHS
$$(n-1)! [z^{n-1}]
\sum_{k=1}^\infty \frac{B_k}{k} (-1)^{n-k} 
\frac{1}{(k-1)!} \left(\log\frac{1}{1-z}\right)^{k-1}
\\ = (n-1)! (-1)^n [z^{n-1}]
\sum_{k=1}^\infty \frac{B_k}{k!} (-1)^{k} 
\left(\log\frac{1}{1-z}\right)^{k-1}
\\ = (n-1)! (-1)^n [z^{n-1}] \left(\log\frac{1}{1-z}\right)^{-1}
\sum_{k=1}^\infty \frac{B_k}{k!} (-1)^{k} 
\left(\log\frac{1}{1-z}\right)^{k}.$$
We  recognize the  exponential  generating function  of the  Bernoulli
numbers which is $$\frac{w}{e^w-1}.$$
The initial term is missing from the sum so we use
$$-1+\frac{w}{e^w-1}.$$
and obtain
$$(n-1)! (-1)^n [z^{n-1}] \left(\log\frac{1}{1-z}\right)^{-1}
\left(-1 - \log\frac{1}{1-z} 
\frac{1}{\exp\left(-\log\frac{1}{1-z}\right)-1}\right)
\\ = (n-1)! (-1)^n [z^{n-1}] \left(\log\frac{1}{1-z}\right)^{-1}
\left(-1 - \log\frac{1}{1-z} 
\frac{1}{(1-z)-1}\right)
\\ = (n-1)! (-1)^n [z^{n-1}]
\left(-\left(\log\frac{1}{1-z}\right)^{-1}
+ \frac{1}{z}\right).$$
Continuing with the RHS we obtain
$$-\frac{1}{n} n! [z^n]
\sum_{k=1}^\infty \frac{1}{k+1} (-1)^{n-k} 
\frac{1}{k!}\left(\log\frac{1}{1-z}\right)^k
\\ = -(n-1)! (-1)^n [z^n]
\sum_{k=1}^\infty \frac{1}{k+1} (-1)^{k} 
\frac{1}{k!}\left(\log\frac{1}{1-z}\right)^k
\\ = -(n-1)! (-1)^n [z^n]
\sum_{k=1}^\infty (-1)^{k} 
\frac{1}{(k+1)!}\left(\log\frac{1}{1-z}\right)^k
\\ = (n-1)! (-1)^n [z^n] \left(\log\frac{1}{1-z}\right)^{-1}
\sum_{k=1}^\infty (-1)^{k+1} 
\frac{1}{(k+1)!}\left(\log\frac{1}{1-z}\right)^{k+1}.$$
This time we recognize
$$-1-w+\exp(w)$$
to get
$$(n-1)! (-1)^n [z^n] \left(\log\frac{1}{1-z}\right)^{-1}
\left(-1+\log\frac{1}{1-z} + 
\exp\left(-\log\frac{1}{1-z}\right)\right)
\\ = (n-1)! (-1)^n [z^n] \left(\log\frac{1}{1-z}\right)^{-1}
\left(-1+\log\frac{1}{1-z} + 1-z\right)
\\ = (n-1)! (-1)^n [z^n] 
\left(1 - z\left(\log\frac{1}{1-z}\right)^{-1}\right).$$
We thus get for $n\gt 1$ for the LHS
$$- (n-1)! (-1)^n [z^{n-1}]
\left(\log\frac{1}{1-z}\right)^{-1}$$
and for the RHS
$$- (n-1)! (-1)^n [z^n] 
z\left(\log\frac{1}{1-z}\right)^{-1}.$$
These two are the same by inspection, QED.
A: Here is a re-write of my first proof with somewhat better aesthetics. 

Suppose we seek to show that
$$\sum_{k=1}^n \frac{B_k}{k} (-1)^{n-k} 
{n-1\brack k-1}
= -\frac{1}{n} 
\sum_{k=1}^n \frac{1}{k+1} (-1)^{n-k} {n\brack k}.$$
Observe that using  the EGF of the Stirling numbers  of the first kind
we have
$${n\brack k}
= \frac{n!}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} 
\frac{1}{k!}\left(\log\frac{1}{1-z}\right)^k \; dz.$$
and using the EGF of the Bernoulli numbers we also have
$$B_k = \frac{k!}{2\pi i} 
\int_{|w|=\epsilon}
\frac{1}{w^{k+1}} \frac{w}{\exp(w)-1} \; dw.$$
The first  of these controls  the range and  we may extend the  sum to
infinity, obtaining for the LHS
$$ \frac{(n-1)! \times (-1)^n}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^n}
\int_{|w|=\epsilon} \frac{1}{\exp(w)-1}
\sum_{k\ge 1} (-1)^k \frac{1}{w^{k}}
\left(\log\frac{1}{1-z}\right)^{k-1}
\; dw \;  dz.$$
This is
$$- \frac{(n-1)! \times (-1)^n}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^n}
\int_{|w|=\epsilon} \frac{1}{\exp(w)-1}
\frac{1}{w} \left(1+\frac{1}{w}
\log\frac{1}{1-z}\right)^{-1}
\; dw \;  dz
\\ = - \frac{(n-1)! \times (-1)^n}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^n}
\int_{|w|=\epsilon} \frac{1}{\exp(w)-1}
\left(w+\log\frac{1}{1-z}\right)^{-1}
\; dw \;  dz.$$
Extracting the residue from the pole at $w=0$ we obtain
$$- \frac{(n-1)! \times (-1)^n}{2\pi i} 
\int_{|z|=\epsilon} \frac{1}{z^n}
\left(\log\frac{1}{1-z}\right)^{-1} \; dz.$$

Next do the RHS, getting
$$-\frac{1}{n} (-1)^n\frac{n!}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} 
\sum_{k\ge 1} (-1)^k 
\frac{1}{(k+1)!}\left(\log\frac{1}{1-z}\right)^k \; dz.$$
The inner term is
$$-\left(\log\frac{1}{1-z}\right)^{-1}
\sum_{k\ge 1} (-1)^{k+1}
\frac{1}{(k+1)!}\left(\log\frac{1}{1-z}\right)^{k+1}
\\ = -\left(\log\frac{1}{1-z}\right)^{-1}
\left(-1+ \log\frac{1}{1-z}
+ \exp\left(-\log\frac{1}{1-z}\right)\right)
\\ = -\left(\log\frac{1}{1-z}\right)^{-1}
\left(-1+ \log\frac{1}{1-z} + 1-z\right)
= -1 + z\left(\log\frac{1}{1-z}\right)^{-1}.$$
This gives for the RHS integral
$$ -\frac{1}{n} \frac{n!\times (-1)^n}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} 
\left(-1 + z\left(\log\frac{1}{1-z}\right)^{-1}\right) \; dz.$$
When $n\ge 1$ this simplifies to
$$-\frac{(n-1)!\times (-1)^n}{2\pi i} 
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} 
z\left(\log\frac{1}{1-z}\right)^{-1} \; dz$$
which is the same as the LHS, QED.
