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?
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.
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.
