Find the limit of $4^n\cdot\binom{2n}{n}/\binom{4n}{2n}$ I am trying to prove that $$f(n)=4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}$$ converges as $n\rightarrow\infty$.
I have already tried to use the fact that, if $n, k \in\mathbb{N}, n\geq k\geq1,$ then $(\frac{n}{k})^k\leq\binom{n}{k}\leq e^k(\frac{n}{k})^k$.  However, this has been to no avail.
Could anybody suggest how I could approach this proof?
 A: This can be done directly, without Stirling. Note that after the simplification of the factorials the sequence becomes
$$ f_n = \frac{2n+2}{2n+1}\frac{2n+4}{2n+3}...\frac{4n}{4n-1}  $$
Taking $\log$ we see that 
$$\log f_n = \log\left(1+\frac{1}{2n+1}\right)+...+\log\left(1+\frac{1}{4n-1}\right)$$
Using the fact that $\log (1+x) \simeq x$ for $x$ close to $0$, we have
$$ \log f_n \simeq \frac{1}{2n+1}+...+\frac{1}{4n-1}$$
Now denote $a_n = \displaystyle \frac{1}{n+1}+...+\frac{1}{2n}$. Using partial sums of a Riemann integral, or the logarithm approximation, we find that $a_n \to \ln 2$.
Now just note that $\displaystyle \frac{1}{2n+1}+...+\frac{1}{4n-1} = a_{2n}-\frac{a_n}{2}$. Therefore 
$$ \log f_n \to \frac{\ln 2}{2}$$
which implies that $f_n \to \sqrt{2}$.
A: Hint: Use the asymptotic properties of the Central Binomial Coefficient to find $\displaystyle\lim_{n\to\infty} f(n)$
$$n\to\infty\implies \binom{2n}{n}\sim \frac{4^n}{\sqrt{\pi n}}\quad\textrm{and}\quad\binom{4n}{2n}\sim\frac{4^{2n}}{\sqrt{2\pi n}}$$
The limit will come out as $\sqrt 2$ proving that the limit converges.
A: Inequality $(9)$ from this answer shows that
$$
\frac{4^n}{\sqrt{\pi(n+\frac13)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi(n+\frac14)}}
$$
Therefore,
$$
\sqrt{\frac{2n+\frac14}{n+\frac13}}\le4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}\le\sqrt{\frac{2n+\frac13}{n+\frac14}}
$$
The Squeeze Theorem says that
$$
\lim_{n\to\infty}4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}=\sqrt2
$$
