Is there a closed form for the sum $\sum_{k=2}^N {N \choose k} \frac{k-1}{k}$?

Question

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.

Answer 1

The Maple command $$ sum(binomial(N, k)*(k-1)/k, k = 2 .. N) $$ outputs $$ 1/4\,{\mbox{$_3$F$_2$}(2,2,-N+2;\,3,3;\,-1)}N \left( N-1 \right) $$ in terms of the hypergeometric function.

Answer 2

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.

Answer 3

In Mathematica Sum[Binomial[n, k] (k - 1)/k, {k, 2, n}] returns:

-1 + 2^n - n HypergeometricPFQ[{1, 1, 1 - n}, {2, 2}, -1]

In traditional form: $-1 + 2^n - n * {}_{3}F_2(1, 1, 1 - n; 2, 2; -1)$