Let $a_k=\frac1{\binom{n}k}$, $b_k=2^{k-n}$. Compute $\sum_{k=1}^n\frac{a_k-b_k}k$

Note that $$\sum_{k=1}^{n}{\frac{1}{k{n \choose k}}} = \frac{1}{n}\sum_{k=0}^{n-1}{\frac{1}{{n-1 \choose k}}} = \frac{1}{n}\sum_{k=0}^{n-1}\frac{\Gamma\left(k+1\right)\Gamma\left(n-k\right)}{\Gamma\left(n\right)} =$$ $$\sum_{k=0}^{n-1}\beta\left(k+1,n-k\right)$$

where $\Gamma$ is the gamma function and $\beta$ is the beta function.

Using the integral form for the beta function, this sum equals $$\sum_{k=0}^{n-1}\int_{0}^{1}t^{k}(1-t)^{n-k-1}dt = \int_{0}^{1}\left(\sum_{k=0}^{n-1}t^{k}(1-t)^{n-k-1}\right)dt =$$ $$\int_{0}^{1}\frac{t^{n} - \left(1-t\right)^{n}}{t - (1-t)}dt = \frac{1}{2^n}\int_{0}^{1}\frac{\left(s+1\right)^n - \left(1-s\right)^n}{s}ds =$$ $$\frac{1}{2^n}\left(\int_{0}^{1}\frac{\left(s+1\right)^n - 1}{s}ds + \int_{0}^{1}\frac{1 - \left(1-s\right)^n}{s}ds\right) =$$ $$\frac{1}{2^n}\sum_{k=0}^{n-1}\left(\int_{0}^{1}\left(1+s\right)^{k}ds + \int_{0}^{1}\left(1-s\right)^{k}ds\right) =$$ $$\frac{1}{2^{n}}\sum_{k=0}^{n-1}\left(\frac{2^{k+1}-1}{k+1} + \frac{1}{k+1}\right) = \frac{1}{2^{n-1}}\sum_{k=0}^{n-1}\frac{2^k}{k+1} = \sum_{k=1}^{n}\frac{2^{k-n}}{k}$$

Which proves that your sum is zero for each $n$.

This argument is taken from this paper.

There is a heavy machinery approach. First, the sum is obviously a hypergeometric sum, and can be fed to Mathematica, which uses the W-Z method to produce:

$$ 2^{-n} \left(2 n \,_3F_2\left(\frac{1}{2},\frac{1}{2}-\frac{n}{2},1-\frac{n}{2}; \frac{3}{2},\fra c{3}{2};1\right)+2^{n+1} \Phi (2,1,n+1)+i \pi \right), $$ where $\Phi$ is Lerch Phi. Now, it is known that the general hypergeometric function with rational parameters can be written as a sum of Lerch Phis (which are basically polylogarithms). See Kelly Roach's amazing paper. Note that Mathematica does NOT know how to simplify this case (though it gets $0$ for particular values).