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?

  • 3
    $\begingroup$ You could use Stirling's Approximation to estimate the central binomial coefficient. $\endgroup$ – 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 $f_n \to \sqrt{2}$.

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    $\begingroup$ (+1) 'without stirling' .. its just another application of partial summation formula on $H_n$. $\endgroup$ – 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.

  • 1
    $\begingroup$ I suppose that you have a typo in the first approximation : the $2$ should not be present in the radical. $\endgroup$ – Claude Leibovici Apr 9 '15 at 12:01
  • $\begingroup$ @ClaudeLeibovici, It was a typo actually. Fixed it now. Thanks for pointing it out :) $\endgroup$ – Prasun Biswas Apr 9 '15 at 12:02
  • $\begingroup$ ...which is the direct consequence of using Stirling's formula $\endgroup$ – Alex Apr 9 '15 at 13:20
  • $\begingroup$ @Alex, yes, I know that. I just wrote it like that since this is also a pretty well-known result. $\endgroup$ – 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 $$ \lim_{n\to\infty}4^n\frac{\dbinom{2n}{n}}{\dbinom{4n}{2n}}=\sqrt2 $$


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