Sum of Binomials Coefficients multiplied by the harmonic numbers

Question

I am interested in solve the next sum:

$\sum_{i=1}^{N} {N \choose i} i^{-G}$

for $G \geq 1$. Some ideas?

Thank you in advance by your help!

Answer 1

This doesn't give a closed form solution, but we can compute a generating function: $$\begin{align*} \sum_{N\geq 0} \left(\sum_{i=1}^N {N\choose i}i^{-G}\right)x^N&=\sum_{i=1}^\infty i^{-G}\sum_{N=i}^{\infty} {N\choose i}x^N\\ &=\sum_{i=1}^{\infty}i^{-G} \frac{x^i}{(1-x)^{i+1}}\\ &=\frac{1}{1-x} \mathrm{Li}_G\left(\frac{x}{1-x}\right), \end{align*} $$ where $\mathrm{Li}_G(t):=\sum_{n\geq 1} n^{-G} t^n$ is the polylogarithm.