Find the closed form of $\sum\limits_{k=0}^n \frac{1}{n\choose k}$: Incomplete Beta function in a combinatoric question

Question

Recently I asked a question about the sum of $\sum_{k=0}^n {n\choose k}^p f\left(k\right)$. Then, I was thinking of the case when $p=-1, f\left(x\right)=1$, which is $\sum_{k=0}^n \dfrac{1}{n\choose k}=\dfrac{1}{n\choose 0}+\dfrac{1}{n\choose 1}+\cdots+\dfrac{1}{n\choose n}$. I have substituted $n=0,1,\cdots,7$ and we get $1,2,\dfrac{5}{2},\dfrac{8}{3},\dfrac{8}{3},\dfrac{13}{5},\dfrac{151}{60},\dfrac{256}{105}$, which seems that there has no obvious sequence. Then, I tried to use Wolfram Alpha to find the answer. It gets no answer when I search $\sum_{k=0}^n \dfrac{1}{n\choose k}$, but when we change $\dfrac{1}{n\choose k}$ into $\dfrac{k!\left(n-k\right)!}{n!}$ and take out the $\dfrac{1}{n!}$, we get $$\dfrac{1}{n!}\sum_{k=0}^n k!\left(n-k\right)!$$ , whose answer can be found by Wolfram Alpha! However, the result is this: $$-\dfrac{i2^{-n-1}\Gamma\left(n+2\right)\left(\pi-i\mathrm{B}_2 \left(n+2,0\right)\right)}{n!}=-2^{-n-1}\left(n+1\right)\left(\mathrm{B}_2 \left(n+2,0\right)+i\pi\right)$$

I was really surprised! How come "$i$" is being here! And what's the "$\mathrm{B}_2$" means? I found that it is called Incomplete Beta Function which has the definition: $$\mathrm{B}_z\left(a,b\right)\equiv\int_0^z x^{a-1} \left(1-x\right)^{b-1}dx$$ So, I tried to find $\mathrm{B}_2 \left(n+2,0\right)$. Substitute $z=2, a=n+2, b=0$ into the definition and it gives: $$\mathrm{B}_2 \left(n+2,0\right)=\int_0^2 \dfrac{x^{n+1}}{1-x}dx$$ However, $\dfrac{x^{n+1}}{1-x}$ is not continuous at $x=1$ and I was stuck at this integral. I thought it may have some relationship with the $i\pi$

Conclusion: There are some questions I want to know, please answer me if you know.

$1$. Explain $\dfrac{1}{n!}\sum_{k=0}^n k!\left(n-k\right)!=-2^{-n-1}\left(n+1\right)\left(\mathrm{B}_2 \left(n+2,0\right)+i\pi\right)$
$2$. Solve the integral $\int_0^2 \dfrac{x^{n+1}}{1-x}dx$ and find the answer

$3$. (Optional) Better way to find the sum $\sum_{k=0}^n \dfrac{1}{n\choose k}$

Thank you!

Answer 1

$$\sum_{k=0}^{n}\binom{n}{k}^{-1} = \frac{1}{\Gamma(n+1)}\sum_{k=0}^{n}\Gamma(k+1)\Gamma(n-k+1)=\frac{\Gamma(n+2)}{\Gamma(n+1)}\sum_{k=0}^{n}B(k+1,n-k+1)$$ using the integral representation for the Beta function this is turned into $$\begin{eqnarray*}(n+1)\int_{0}^{1}\sum_{k=0}^{n}x^{k}(1-x)^{n-k}\,dx &=&(n+1)\int_{0}^{1}\frac{x^{n+1}-(1-x)^{n+1}}{2x-1}\,dx\\&=&\frac{n+1}{2}\int_{-1}^{1}\frac{\left(\frac{1+z}{2}\right)^{n+1}-\left(\frac{1-z}{2}\right)^{n+1}}{z}\,dz\end{eqnarray*}$$ or $$\begin{eqnarray*}\frac{n+1}{2^{n+2}}\int_{-1}^{1}\left[(1+z)^{n+1}-(1-z)^{n+1}\right]\frac{dz}{z}&=&\frac{n+1}{2^{n}}\int_{0}^{1}\sum_{k\leq n/2}\binom{n+1}{2k+1}z^{2k}\,dz\\&=&\frac{n+1}{2^n}\sum_{k\leq n/2}\binom{n+1}{2k+1}\frac{1}{2k+1}.\end{eqnarray*}$$