One of those...

$\large A = \frac{\binom{2n + 1}{n + 1}}{\binom{2n - 1}{n}} = \frac{\frac{(2n + 1)!}{(n+1)! \cdot n!}}{\frac{(2n - 1)!}{n! \cdot (n - 1)!}} = \frac{\frac{(2n + 1)!}{(n+1)!}}{\frac{(2n - 1)!}{(n - 1)!}} = \frac{(2n + 1) \cdot 2n}{(n + 1) \cdot n} = \frac{2(2n +1)}{n + 1}$ Now, substituing $n = 10^9$ and using a calculator $\large \frac{\binom{2\cdot 10^9 + 1}{10^9 + 1}}{\binom{2\cdot 10^9 - 1}{10^9}} = \frac{2(2 \cdot 10^9 + 1)}{10^9 + 1} \approx 3.999999$

Note.- Inspiration: this exercise

$\begin{aligned} A & = \frac {2n+1 \choose n+1}{2n-1 \choose n} \\ & = \frac {(2n+1)!}{n!(n+1)!} \cdot \frac {n!(n-1)!}{(2n-1)!} \\ & = \frac {(2n+1)!}{(n+1)!} \cdot \frac {(n-1)!}{(2n-1)!} \\ & = \frac {2n(2n+1)}{n(n+1)} \\ & = \frac {2(2n+1)}{n+1} \\ & = \frac {4n+4-2}{n+1} \\ & = 4 - \frac 2{n+1} \end{aligned}$

For $n=10^9$ , $A = 4 - \dfrac 2{10^9+1} \approx 4 - 0.000000002 = 3.999999998 \approx \boxed{4.000000}$ .