<s>[INST] The sum $\sum^{n}_{k=0} \frac{\binom{n}{k}}{\binom {2n-1}{k}}x^k$ I was trying to evaluate:
$$\sum_{k=0}^{n} \frac{\binom{n}{k}}{\binom {2n-1}{k}}x^k$$
And in particular, when $x=1$.
I already have $2$ proofs for the $x=1$ case, (it is $2$) by counting subsets of a $2n-1$ size set and by splitting the term into a telescoping sum. I was looking for a solution using only generating functions. 
 [/INST] (I started an answer some hours ago, but had to stop, well, it is time to finish and submit.) Let $y=1/x$, (if $x$ is not zero,) and let us compute closer for some integer $n>0$
$$
\begin{aligned}
\frac{\binom nk}{\binom {2n-1}k}
&=\frac{n!}{k!(n-k)!}\cdot\frac{k!(2n-k-1)!}{(2n-1)!}
\\
&=\frac{n!(n-1)!}{(2n-1)!}\cdot\frac{(2n-k-1)!}{(n-k)!(n-1)!}
\\
&=\binom{2n-1}{n}^{-1}\cdot\binom{2n-k-1}{n-1}\ .
\\ %[3mm]
&\qquad\text{ This implies for the sum to be calculated:}
\\
f(x)
&:=
\sum_{0\le k\le n} 
\frac{\binom nk}{\binom {2n-1}{k}}x^k
=
\binom{2n-1}{n}^{-1}
\sum_{0\le k\le n} 
\binom{2n-k-1}{n-1} x^k
\\
&=
x^n
\binom{2n-1}{n}^{-1}
\sum_{0\le k\le n} 
\binom{n-1+(n-k)}{n-1} y^{n-k}\qquad\text{ (Recall $y=1/x$)}
\\
&\qquad\text{ and let us use in the summation the variable $j=n-k$}
\\
&=
x^n
\binom{2n-1}{n}^{-1}
\sum_{n\ge j\ge 0} 
\binom{n-1+j}{n-1} y^j\ .
\end{aligned}
$$
Now it is time to set $x=y=1$, and recall the formula:
$$
\binom KK +
\binom {K+1}K +
\binom {K+2}K +
\dots +
\binom NK 
=
\binom{N+1}{K+1}\ .\qquad(*)
$$
This gives
$$
f(1) = 
\binom{2n-1}{n}^{-1}
\sum_{n\ge j\ge 0} 
\binom{n-1+j}{n-1} 
=
\binom{2n-1}{n}^{-1}
\binom{(2n-1)+1}{(n-1)+1}
=
\binom{2n-1}{n}^{-1}
\binom{2n}{n}
=2\ .
$$
(Because of $\binom{2n}n=\frac{2n}n\binom{2n-1}{n-1}=2\binom{2n-1}{n}$.)
For a more general formula of $f(x)$ containing the variable $x$, we would need a corresponding relation $(*)$ containing $x$.
</s>