While playing around with the summation presented in a recent answer, I noticed that the following seems to be empirically true (up to $n=100$ via Mathematica):
$$\sum_{k=1}^{n} \frac{1}{k}\binom{n}{k}^{-1}=\sum_{k=1}^{n} \frac{1}{k}\frac{1}{2^{n-k}}$$
I'm not so familiar with summations involving reciprocals of binomial coefficients, and I haven't (yet) found an earlier question which covers this. Judging from earlier answers to such questions, they often revolve around the integral representation $$\binom{n}{k}^{-1}=k\int_0^1 (1-t)^{k-1}t^{n-k}dt$$ as a means to clear the summation over $k$. This initially seems attractive, particularly the role of $1/k$ to clear the coefficient. Here, though, the summation over $k$ is the entire point, so I don't see how that helps. Alternatively, we can straightforwardly pass the RHS to a generating function:
\begin{align}
\sum_{n=1}^\infty \sum_{k=1}^n \frac{2^{k-n}}{k}x^n
&=\sum_{k=1}^\infty \frac{2^k}{k}\sum_{n=k}^\infty \left(\frac{x}{2}\right)^n\\
&=\sum_{k=1}^\infty \frac{2^k}{k}\frac{(x/2)^k}{1-x/2}\\
&=\frac{1}{1-x/2}\sum_{k=1}^\infty \frac{x^k}{k}=\frac{\ln(1-x)}{1-x/2}
\end{align}
But then it's not obvious how to relate this to $\binom{n}{k}^{-1}$.
Best Answer
We seek to show that
$$\sum_{k=1}^n \frac{1}{k} {n\choose k}^{-1} = \sum_{k=1}^n \frac{1}{k} \frac{1}{2^{n-k}}.$$
Recall from MSE 4316307 the following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1} = [v^n] \log\frac{1}{1-v} (v-1)^{n-k}.$$
We get for the LHS
$$[v^n] \log\frac{1}{1-v} (v-1)^{n-1} \sum_{k=1}^n (v-1)^{-(k-1)} \\ = [v^n] \log\frac{1}{1-v} (v-1)^{n-1} \frac{1-1/(v-1)^n}{1-1/(v-1)} \\ = [v^n] \log\frac{1}{1-v} (v-1)^n \frac{1-1/(v-1)^n}{v-2}.$$
We get two pieces, the simple one is
$$-[v^n] \log\frac{1}{1-v} \frac{1}{v-2} \\ = \frac{1}{2} [v^n] \log\frac{1}{1-v} \frac{1}{1-v/2} = \frac{1}{2} \sum_{k=1}^n \frac{1}{k} \frac{1}{2^{n-k}}.$$
The second piece is
$$\;\underset{v}{\mathrm{res}}\; \frac{1}{v^{n+1}} \log\frac{1}{1-v} (v-1)^n \frac{1}{v-2}.$$
Now put $v/(v-1) = w$ so that $v=w/(w-1)$ and $dv = -1/(w-1)^2 \; dw$ to get
$$- \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} (w-1) \log\frac{1}{1-w/(w-1)} \frac{1}{w/(w-1)-2} \frac{1}{(w-1)^2} \\ = - \;\underset{w}{\mathrm{res}}\; \frac{1}{w^{n+1}} \log(1-w) \frac{1}{2-w} = \frac{1}{2} \sum_{k=1}^n \frac{1}{k} \frac{1}{2^{n-k}}.$$
Add the two pieces to obtain
$$\sum_{k=1}^n \frac{1}{k} \frac{1}{2^{n-k}}$$
as claimed.