Is there a closed form for the sum $\sum_{k=2}^N {N \choose k} \frac{k-1}{k}$? I am interested in finding a closed form for the sum $\sum_{k=2}^N {N \choose k} \frac{k-1}{k}$. Does anyone know if there is some Binomial identity that might be helpful here?
Thank you.

You might consider expanding $$(1+x)^N - \int \frac{(1+x)^N}{x} dx$$ and then letting $x=1$.  This will need some slight adjustment as it has a few extra terms compared with your sum.
The problem is the integral as it involves a hypergeometric function. 
