Nice solution. This problem is I generalization of problem 158 (
http://people.missouristate.edu/lesreid/Adv158.html) I sent to Missouri State University Problem Corner. Here is my approach:
Using the notation \(\Gamma(k+1)=k!\) for \(k\in\mathbb{N}\cup\{0\}\), where \(\Gamma\) is the Gamma function, we will show that
$$\sum_{k\geq1}(-1)^{k-1}\frac{1}{k\cdot(k+1)\cdots(k+m)}=\frac{2^m}{\Gamma(m+1)}\left(\ln2-\sum_{k=1}^{m}\frac{\left(1/2\right)^k}{k}\right),\hspace{10ex}m\in\mathbb{N}.$$
Using the identity \(\Gamma(k+1)=k\Gamma(k),\quad k>0\), by a direct calculation we see that
\begin{equation}\label{1}\frac{\Gamma(k)}{\Gamma(k+m+1)}=\frac{1}{m}\left(\frac{\Gamma(k)}{\Gamma(k+m)}-\frac{\Gamma(k+1)}{\Gamma(k+m+1)}\right),\hspace{10ex}m\in\mathbb{N}.\end{equation}
Now setting \(A_{m,n}:=\sum_{k=1}^{n}(-1)^{k-1}\frac{1}{k\cdot(k+1)\cdots(k+m)},\quad m\in\mathbb{N}\cup\{0\}\), on account of \((1)\) we have:
\(\begin{align}A_{m,n}&=\sum_{k=1}^{n}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m+1)}\notag \\ &=\frac{1}{m}\sum_{k=1}^{n}(-1)^{k-1}\left(\frac{\Gamma(k)}{\Gamma(k+m)}-\frac{\Gamma(k+1)}{\Gamma(k+m+1)}\right)\notag \\&=\frac{1}{m}\left(\frac{1}{\Gamma(m+1)}+(-1)^n\frac{\Gamma(n+1)}{\Gamma(n+m+1)}+2\sum_{k=2}^{n}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m)}\right)\notag\\&=\frac{1}{m}\left(\frac{1}{\Gamma(m+1)}+(-1)^n\frac{\Gamma(n+1)}{\Gamma(n+m+1)}+2\left(A_{m-1,n}-\frac{1}{\Gamma(m+1)}\right)\right),\notag\end{align}\)
so \(\displaystyle A_{m,n}=\frac{1}{m}\left(-\frac{1}{\Gamma(m+1)}+(-1)^{n}\frac{\Gamma(n+1)}{\Gamma(n+m+1)}+2A_{m-1,n}\right),\) and, since for \(m\in\mathbb{N}\cup\{0\}\) by Dirichlet's criterion \(\displaystyle\sum_{k\geq1}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m+1)}\) converges and for \(m\in\mathbb{N}\) we have \(\displaystyle(-1)^{n}\frac{\Gamma(n+1)}{\Gamma(n+m+1)}\to0\), setting \(\displaystyle A_m:=\sum_{k\geq1}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m+1)}\) and letting \(n\to+\infty\) we have
\(\begin{equation}\label{2}\displaystyle A_m=-\frac{1}{m\Gamma(m+1)}+\frac{2}{m}A_{m-1},\hspace{3ex}m\in\mathbb{N}.\end{equation}\)
Since \(A_{0}=\ln2\), with a simple inductive argument, \((2)\) yields
\(\displaystyle A_m=\frac{2^m}{\Gamma(m+1)}\left(\ln2-\sum_{k=1}^{m}\frac{\left(1/2\right)^k}{k}\right),\hspace{10ex}m\in\mathbb{N}\cup\{0\}.\)