# 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?

• You could use Stirling's Approximation to estimate the central binomial coefficient. – r9m Apr 9 '15 at 11:47

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 $f_n \to \sqrt{2}$.

• (+1) 'without stirling' .. its just another application of partial summation formula on $H_n$. – r9m Apr 9 '15 at 12:49

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.

• I suppose that you have a typo in the first approximation : the $2$ should not be present in the radical. – Claude Leibovici Apr 9 '15 at 12:01
• @ClaudeLeibovici, It was a typo actually. Fixed it now. Thanks for pointing it out :) – Prasun Biswas Apr 9 '15 at 12:02
• ...which is the direct consequence of using Stirling's formula – Alex Apr 9 '15 at 13:20
• @Alex, yes, I know that. I just wrote it like that since this is also a pretty well-known result. – Prasun Biswas Apr 9 '15 at 13:21

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 $$\lim_{n\to\infty}4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}=\sqrt2$$