Is there a counting argument for the following identity? $$\sum_{k=1}^{n}\sum_{j=0}^{k-1}(-1)^{j}\binom{n}{j}\frac{1}{k} = \frac{1}{n}$$ where $n$ is a positive integer
NOTE: I do NOT need algebraic proofs. I have already proved the identity. I only want to know if there is a counting argument for the identity.
I stumbled across this identity while trying to prove $$\sum_{k=1}^{\infty}\frac{1}{k(k+1)...(k+n)}=\frac{1}{n\cdot n!}$$ I am aware of the counting argument for $\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}=0$, which involves producing a bijection from the set of all even subsets of $\{1,2,3,...n\}$ to the set of all odd subsets. However, I am not sure how one would put the $\frac{1}{k}$ together with the $(-1)^{j}\binom{n}{j}$.
You can do it as a telescoping series.
Let $p_n(x) =\prod_{k=0}^{n-1} (x+k) $.
Then
$\begin{array}\\ \dfrac1{p_n(x)}-\dfrac1{p_n(x+1)} &=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)}\\ &=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\ &=\dfrac{(x+n)-x}{\prod_{k=0}^{n} (x+k)}\\ &=\dfrac{n}{p_{n+1}(x)}\\ \end{array} $
so that
$\dfrac{1}{p_{n+1}(x)} =\dfrac1{n}(\dfrac1{p_n(x)}-\dfrac1{p_n(x+1)}) $
and the series telescopes.
Also note that
$\begin{array}\\ p_n(x+1)-p_n(x) &=\prod_{k=0}^{n-1} (x+1+k)-\prod_{k=0}^{n-1} (x+k)\\ &=\prod_{k=1}^{n} (x+k)-\prod_{k=0}^{n-1} (x+k)\\ &=\prod_{k=1}^{n-1} (x+k)(x+n-x)\\ &=n\prod_{k=0}^{n-2} (x+1+k)\\ &=np_{n-1}(x+1)\\ \end{array} $
so that $\sum_k p_n(x+k) $ also telescopes.