Combinatorially showing $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$

Question

I am trying to show that $\lim_{n\to \infty}{\frac{2n\choose n}{4^n}}=0$. I found that using stirling's approximation, I can get: $$ \lim_{n\to \infty}{\frac{2n\choose n}{4^n}}= \lim_{n\to \infty}{\frac{(2n)!}{n!^2*2^{2n}}}= \lim_{n\to \infty}{\frac{(\frac{2n}{e})^{2n}\sqrt{2\pi (2n)}}{((\frac{n}{e})^n\sqrt{2\pi n})^2*2^{2n}}} =\lim_{n\to \infty}{\frac{2\sqrt{\pi n}}{2\pi n}} =0 $$

But this seems inelegant where there should be a more elegant, combinatorial proof. Is there an easier way?

Answer 1

Not a combinatorial approach, but a fun way of doing it is showing that $$\frac{\binom{2n}{n}}{4^n}=\frac{1}{2\pi}\int_0^{2\pi} \cos^{2n} x\; dx$$

Basically, writing $\cos x = \frac12\left(e^{ix}+e^{-ix}\right)$. So $\cos^{2n} x$ has constant term $\binom{2n}{n}/4^n$.

Since $\cos^{2n} x\to 0$ for almost all $x$, it is pretty easy to show that the above integral tends to zero as $n\to\infty$.
Specifically, on the intervals $(\varepsilon,\pi-\varepsilon)\cup(\pi+\varepsilon,2\pi-\varepsilon)$, $\cos^{2n}x\to 0$ uniformly.

Answer 2

Here’s a non-combinatorial argument that avoids Stirling’s approximation.

$$\begin{align*} p_n\triangleq\frac{\binom{2n}n}{4^n}&=\frac{(2n)!}{(2^nn!)^2}\\\\ &=\frac{(2n)!(2n-1)!!^2}{(2n)!^2}\\\\ &=\frac{(2n-1)!!^2}{(2n)!}\\\\ &=\prod_{k=1}^n\frac{2k-1}{2k}\\\\ &=\prod_{k=1}^n\left(1-\frac1{2k}\right)\;, \end{align*}$$

so

$$\ln p_n=\sum_{k=1}^n\left(-\sum_{m\ge 1}\frac1{m(2k)^m}\right)\le-\sum_{k=1}^n\frac1{2k}=-\frac12H_n\;,$$

where $H_n$ is the $n$-th harmonic number. The harmonic series diverges, so $\lim_{n\to\infty}\ln p_n=-\infty$, and therefore $\lim_{n\to\infty}p_n=0$.

Note: The double factorial $(2n-1)!!$ is the product of the odd positive integers not exceeding $2n-1$:

$$(2n-1)!!=\prod_{k=1}^n(2k-1)=\frac{(2n)!}{2^nn!}\;.$$

Answer 3

Another was is to use the generating function of the central binomial coefficients, that is $$\frac{1}{\sqrt{1-x}}=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}}x^{n}.$$ We will make use of the fact that $x_{n}=\binom{2n}{n}\frac{1}{4^{n}}$ is a monotically decreasing sequence to make sure there are no oscillations. It is monotonic since $$x_{n+1}=\binom{2n}{n}\frac{1}{4^{n}}\frac{(2n+2)(2n+1)}{4\left(n+1\right)^{2}}=x_{n}\left(1-\frac{1}{2n+2}\right).$$ Taking the indefinite integral of both sides of the generating function, we have that for $|x|<1,$ $$2-2\sqrt{1-x}=\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}}\frac{x^{n+1}}{n+1}.$$ Now, the left hand side converges as $x\rightarrow1$ from the left, and so, since the series has strictly positive coefficients, the right hand side must converge as well. This implies that $$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}}\frac{1}{n+1}$$ is a convergent series, and since the coefficients $\binom{2n}{n}\frac{1}{4^{n}}$ are monotonically decreasing, the divergence of the harmonic series implies that we must have $$\binom{2n}{n}\frac{1}{4^{n}}\rightarrow0.$$

Added Details: Based on the comments, I thought I would give a precise proof of why the series must converge, without appealing to Littlewood's Tauberian theorem. The proof only requires that the coefficients of the power series, $a_n$, are nonnegative. Let $a_n\geq0$, $s_m=\sum_{n=0}^m a_n$ be the partial sums, and suppose that $\sum_{n=0}^{\infty}a_{n}x^{n}\leq C$ for all $x\in(0,1),$ for some positive constant $C$. Then for $x\in(0,1),$ $$s_{m}=\sum_{n=0}^{m}a_{n}\left(1-x^{n}\right)+\sum_{n=0}^{m}a_{n}x^{n}$$ $$\leq(1-x)\sum_{n=0}^{m}a_{n}\left(1+x+\cdots+x^{n-1}\right)+C$$

$$\leq(1-x)\sum_{n=0}^{m}na_{n}+C.$$ Taking $x$ sufficiently close to $1$, we can make $(1-x)\sum_{n=0}^{m}na_{n}\leq 1$, and so it follows that $s_{m}\leq C+1.$ This means that $s_m$ is monotonically increasing and bounded, its limit, $\sum_{n=0}^\infty a_n$ must converge. In particular, our series $$\sum_{n=0}^{\infty}\binom{2n}{n}\frac{1}{4^{n}(n+1)},$$ is convergent.

Answer 4

Here is what I think "a combinatorial approach:" Note that $\binom{2n}{n}$ is the number of $n$ element subsets of a $2n$ element set. Consider the quantities $\binom{2n}{n+i},\,i=1,\ldots,k$ for some small $k$ (relative to $n$). We have the relationship $\binom{2n}{n+1} = \frac{n}{n+1}\binom{2n}{n}$, and in general, $\binom{2n}{n+k} = \frac{n-k+1}{n+k}\binom{2n}{n+k-1}$. Now for any $k$, we can choose $n$ so large that $\frac{n-k+1}{n+k}$ is at least $1-\frac{1}{2k}$. This gives us the estimate $\binom{2n}{n+k} \geq (1-\frac{1}{2k})\binom{2n}{n+k-1} \geq \cdots \geq (1-\frac{1}{2k})^k\binom{2n}{n} \geq \frac{1}{2}\binom{2n}{n}$. Therefore, for any $i\in\{1,\ldots,k\}$ and when $n$ is large enough, the number of $n+i$ element subsets of a $2n$ element set is at least $\frac{1}{2}\binom{2n}{n}$. Hence, $\frac{k}{2}\binom{2n}{n} \leq 4^n$. Now, let $k\rightarrow\infty$.

Answer 5

Notice that $$ \begin{align} (n+2)(2n+1)^2 &=4n^3+12n^2+9n+2\\ &\le4n^3+12n^2+12n+4\\ &=(n+1)(2n+2)^2\tag1 \end{align} $$ which shows that $$ \left(\frac{2n+1}{2n+2}\right)^2\le\frac{n+1}{n+2}\tag2 $$

Since $$ \frac{\frac1{4^{n+1}}\binom{2n+2}{n+1}}{\frac1{4^n}\binom{2n}{n}}=\frac{2n+1}{2n+2}\le\sqrt{\frac{n+1}{n+2}}\tag3 $$ we can take products to get $$ \begin{align} \frac1{4^n}\binom{2n}{n} &=\prod_{k=0}^{n-1}\frac{\frac1{4^{k+1}}\binom{2k+2}{k+1}}{\frac1{4^k}\binom{2k}{k}}\tag{4a}\\ &\le\prod_{k=0}^{n-1}\sqrt{\frac{k+1}{k+2}}\tag{4b}\\[6pt] &=\frac1{\sqrt{n+1}}\tag{4c} \end{align} $$ which shows that $$ \lim_{n\to\infty}\frac1{4^n}\binom{2n}{n}=0\tag5 $$

Answer 6

Speaking of combinatorial proofs, $$ \sum_{k=0}^n\frac{\binom{2k}{k}}{2^{2k}}\frac{\binom{2(n-k)}{n-k}}{2^{2(n-k)}}=1. $$ The key step of a combinatorial proof is that $p_k=\binom{2k}k2^{-2k}$ is the probability that random path with $2k$ steps $(1,1)$ and $(1,-1)$ doesn't touch $x$-axis after the start. So now we see that sequence $p_k$ is monotonic. But then the first identity obviously implies $p_k\to0$ (if $\forall k\;p_k\gt\epsilon$, $1=\sum p_kp_{N-k}\gt N\epsilon\gt1$)

Answer 7

A probabilistic approach that may content you. Let $p_n={2n\choose n}/4^n$; $p_n$ is the probability that a random walker with equally likely steps $-1$ and $+1$ is found at the origin after $2n$ steps. We can prove that $p_n\to 0$ using the fact that a binomial distribution "flattens out" as $n$ gets large.

Lemma. For all $k\in\mathbb{N}$ , $$\lim_{n\to\infty}\frac{2n\choose n+k}{2n\choose n} = 1.$$

Proof. \begin{align*} \lim_{n\to\infty}\frac{2n\choose n+k}{2n\choose n}&=\lim_{n\to\infty}\frac{(n-k)!(n+k)!}{(n!)^2}\\ &=\lim_{n\to\infty}\frac{(n+k)(n+k-1)...(n+1)}{n(n-1)...(n-k+1)}\\ &=1 \end{align*}

Let us now suppose BWOC that $p_n$ does not tend to $0$. Then there must exist some $\epsilon>0$ such that the set $S=\{n\in\mathbb{N}:p_n>\epsilon\}$ is unbounded. Let $N\in\mathbb{N}$ be such that $\frac{2}{N}<\epsilon$. By our lemma we can find $n_1,n_2,...,n_N$ such that \begin{align*} n\geq n_j&\implies \left|\frac{2n\choose n+j}{2n\choose n}-1\right|<\frac12\\ &\implies {2n\choose n+j}>\frac{2n\choose n}{2}\\ &\implies \frac{2n\choose n+j}{4^n}>\frac{p_n}{2} \end{align*} for $1\leq j\leq N$. Now, given the unboundedness of $S$, we may find $m\in S$ such that $m\geq \max(n_1,...,n_N)$. Then, observe that for every $1\leq j\leq N$ \begin{align*} \frac{2m\choose m+j}{4^m}&>\frac{p_m}{2}\\ &>\frac{\epsilon}2>\frac1{N}. \end{align*} Therefore $$\sum_{j=1}^N\frac{2m\choose m+j}{4^m}>N\frac1{N}=1.$$ But this contradicts that $$\sum_{j=1}^N\frac{2m\choose m+j}{4^m}<\sum_{j=0}^{2m}\frac{2m\choose j}{4^m}=1.$$ Therefore $\lim_{n\to\infty}p_n=0$.