Proving that $\sum_{k=0}^n\frac{1}{n\choose k}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}$ I want to prove for any positive integer $n$, the following equation holds:
$$\sum_{k=0}^n\frac{1}{n\choose k}=\frac{n+1}{2^{n+1}}\sum_{k=1}^{n+1}\frac{2^k}{k}$$
I tried to expand $2^k$ as $\sum_{i=0}^k{k\choose i}$ and interchange summation, also tried let $f(x)=\sum_{k=1}^{n+1}\frac{x^k}{k}$ and compute $f'(x)$. But it seems I'm not on the right path.

Note that successive terms in the sum on the left turn out have a simple expression for their sum: when $k \ne 0$, we have
$$\frac{1}{\binom{n}{k-1}} + \frac1{\binom{n}{k}} = \frac{\binom{n}{k}+\binom{n}{k-1}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\binom{n+1}{k}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\frac{n+1}{k}\binom{n}{k-1}}{\binom{n}{k-1}\binom{n}{k}} = \frac{\frac{n+1}{k}}{\binom{n}{k}}$$
where we've used Pascal's rule and the "absorption" identity that $\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}$ for $k \neq 0$. Applying absorption again further gives the above expression to be equal to
$$\frac{\frac{n+1}{k}}{\binom{n}{k}} = \frac{\frac{n+1}{k}}{\frac{n}{k}\binom{n-1}{k-1}} = \frac{n+1}{n} \frac{1}{\binom{n-1}{k-1}}$$
This gives a strategy for evaluating the sum on the left:
$$\begin{align}
2\sum_{k=0}^n \frac{1}{\binom{n}{k}} 
&= \sum_{k=1}^{n} \left( \frac{1}{\binom{n}{k-1}} + \frac1{\binom{n}{k}}\right) + 2 \\
&= 2 + \frac{n+1}{n} \sum_{k=1}^{n} \frac{1}{\binom{n-1}{k-1}}
\end{align}$$
or, calling the left-hand-side sum as $L_n = \sum_{k=0}^n \frac{1}{\binom{n}{k}}$, we have
$$2 L_n = 2 + \frac{n+1}{n} L_{n-1}$$
$$L_n = \frac{n+1}{2n}L_{n-1} + 1$$
Calling the right-hand-side term $R_n$, we have
$$\begin{align}
\frac{2^{n+1}}{n+1} R_n &= \sum_{k=1}^{n+1}\frac{2^k}{k} \\
&= \frac{2^n}{n} R_{n-1} + \frac{2^{n+1}}{n+1}
\end{align}$$
thus
$$R_n = \frac{n+1}{2n}R_{n-1} + 1$$
and both the LHS $L_n$ and RHS $R_n$ satisfy the same recurrence and have the same initial values (check for $n=1$, say), so they are equal.
