Reciprocal binomial coefficient polynomial evaluation

Solution 1:

Too long for a comment.

@alduan gave a link to this very interesting paper but I am afraid that there is a mistake somewhere.

I think that it should be

$$\color{blue}{g(x,n)=\sum_{i=0}^n \frac{x^i}{{ n \choose i}}}= \color{red}{\frac 1 x} \color{blue}{\frac {n+1} {\left(1+\frac{1}{x}\right)^{n+1}}\sum_{k=1}^{n+1}\frac 1 k(x^k+1)\left(1+\frac{1}{x}\right)^{k-1}}$$ Otherwise, the first term of the expansion would be $x$ instead of $1$.

For sure, for $x=1$, the result is correct but this not the case for $x\neq1$ (easy to check).

For example, as written is the paper $$g(x,1)=\frac{2 \left(\frac{1}{2} \left(\frac{1}{x}+1\right) \left(x^2+1\right)+x+1\right)}{\left(\frac{1}{x}+1\right)^2}=x(1+x)$$

I sent an e-mail to Prof. Mansour for clarification.

Edit

We exchanged a few e-mails and, effectively, in the paper at ArXiv, one term is missing. The front term must be divided by $(a+b)$.

Have a look at the final version of the paper $$\sum_{k=0}^na^kb^{n-k}\binom nk^{-1}=\frac{n+1}{\color{red}{(a+b)}(\frac1a+\frac1b)^{n+1}}\sum_{k=1}^{n+1}\frac{(a^k+b^k)(\frac1a+\frac1b)^{k-1}}k$$