Reciprocal binomial coefficient polynomial evaluation The conventional binomial coefficient can be obtained via
$$
f(x, n) = (1+x)^n = \sum_{i=0}^n { n \choose i} x^i
$$
And the function $f$ can be every efficiently performed on evaluation.
I'm interested in evaluating value for a function $g(x, n)$ very similar to it
$$
g(x, n) := \sum_{i=0}^n \frac{1}{{ n \choose i}} x^i
$$
With some fixed $n$, how can I convert $g(x, n)$ into a form so it can be evaluated efficiently?
 A: 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
$$\sum_{k=0}^na^kb^{n-k}\binom nk^{-1}=\frac{n+1}{\color{red}{(a+b)}(\frac1a+\frac1b)^{n+1}}\sum_{k=1}^{n+1}\frac{(a^k+b^k)(\frac1a+\frac1b)^{k-1}}k$$
