Compute $\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}$

Question

I want to calculate $\sum_{k=0}^{n}\frac{1}{\binom{n}{k}}$.

No idea in my mind. Any help?

Context

I want to calculate the expected value of bits per symbols in adaptive arithmetic coding when the number of symbols goes to infinity.

Answer 1

$$\sum^{n}_{k=0}\frac{1}{n\choose k}=\sum^{n}_{k=0}\frac{(n-k)!k!}{n!}=\sum^{n}_{k=0}\frac{\Gamma(n-k+1)\Gamma(k+1)}{\Gamma(n+1)}=\sum^{n}_{k=0}(n+1)B(n-k+1,k+1)$$ The last summation is in terms of Euler's Beta function. Beta function has the following integral represantion: $$B(x,y)=\int^{1}_{0}u^{x-1}(1-u)^{y-1}\,du$$ for $\Re(x),\Re(y)>0$. Hence we could rewrite the last summation as follows \begin{align}\sum^{n}_{k=0}(n+1)B(n-k+1,k+1)&=(n+1)\sum^{n}_{k=0}B(n-k+1,k+1)\\&=(n+1)\sum^{n}_{k=0}\int^{1}_{0}u^{n-k}(1-u)^{k}\,du\\& =(n+1)\int^{1}_{0}\{\sum^{n}_{k=0}u^{n-k}(1-u)^{k})\},du\\& =(n+1)\int^{1}_{0}u^n\{\sum^{n}_{k=0}(\frac{1}{u}-1)^{k})\}\,du\\ &=(n+1)\int^{1}_{0}u^n\frac{(\frac{1}{u}-1)^{n+1}-1}{\frac{1}{u}-2}\,du\\ &=(n+1)\int^{1}_{0}\frac{(1-u)^{n+1}-u^{n+1}}{1-2u}\,du\\ &=-\frac{2(n+1)}{n+2}\cdot_2F_1(1,n+2,n+3,2)\end{align} where the last expression is the value of the integral in terms of hypergeometric function.

Answer 2

Very nice!

I have the same idea the final result is: $$F(1,1,-n,0)=\left(\frac{G(-n)G(-n-2)}{G(-n-1)^2}\right)=\left(\frac{\pi}{-\sin(n+1)\pi}\right)n!\left(\frac{\pi}{(-\sin(n+3)\pi}\right)\\(n+1)!\left(\frac{\pi}{(-\sin(n+2)\pi}\right)(n+1)!\left(\frac{-2(n+1)}{(n+2)}\right)$$

(There is a line break between the two parts of the equation, for those wondering)