Computations with Maple suggest the following binomial identity
\begin{equation*} \forall{p,j}: \sum_{k=j+1}^{p+1} (-1)^j \dfrac{1}{k}\binom{k-1}{j} = \sum_{k=j+1}^{p+1} (-1)^{k-1} \dfrac{1}{k}\binom{p+1}{k} \end{equation*}
which is insofar remarkable as on the LHS the summation runs over the upper index of the binomial coefficients whereas on the RHS it runs over the lower index. Any clues for the proof?
$\newcommand{\bi}{\binom} $Denote the left- and right-hand sides of the identity by $l_{j,p}$ and $r_{j,p}$, respectively. Note that \begin{equation*} l_{j,p+1}-l_{j,p}=\frac{(-1)^j}{p+2}\,\bi{p+1}j \end{equation*} and \begin{equation*} r_{j,p+1}-r_{j,p}=\frac{(-1)^{p+1}}{p+2}+s_{j,p}, \end{equation*} where \begin{equation*} s_{j,p}:=\sum_{k=j+1}^{p+1}\frac{(-1)^{k-1}}k\Big(\bi{p+2}k-\bi{p+1}k\Big) = \sum_{k=j+1}^{p+1}\frac{(-1)^{k-1}}k \bi{p+1}{k-1}. \end{equation*} By induction on $p$, it is enough to show that $l_{j,p+1}-l_{j,p}=r_{j,p+1}-r_{j,p}$, which can be rewritten as \begin{equation*} s_{j,p}=u_{j,p}:=\frac{(-1)^j}{p+2}\,\bi{p+1}j-\frac{(-1)^{p+1}}{p+2}. \tag{1}\label{1} \end{equation*}
We have \begin{equation*} s_{j,p}-s_{j+1,p}=\frac{(-1)^j}{j+1} \bi{p+1}j \end{equation*} and \begin{equation*} u_{j,p}-u_{j+1,p}=\frac{(-1)^j}{p+2} \Big(\bi{p+1}j+\bi{p+1}{j+1}\Big) \\ =\frac{(-1)^j}{p+2}\bi{p+2}{j+1} =\frac{(-1)^j}{j+1} \bi{p+1}j=s_{j,p}-s_{j+1,p}. \end{equation*} So, \eqref{1} follows by induction on $j$. $\quad\Box$
