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.

I don't think that there exist a closed form since:
$$\sum_{k=2}^{n}\binom{n}{k}\frac{k-1}{k}=\sum_{k=1}^{n}\binom{n}{k}\frac{k-1}{k}$$$$=\color{blue}{\sum_{k=1}^{n}\binom{n}{k}}-\sum_{k=1}^{n}\binom{n}{k}\frac{1}{k}$$$$=\color{blue}{2^{n}-1}-\sum_{k=1}^{n}\frac{2^{k}}{k}+H_{n}$$
And we have a problem with $\sum_{k=1}^{n}\frac{2^{k}}{k}$.
