Binomial series problem $\lim_{n\to\infty}\left[\sum_{i,j}{2i\choose i}{2j\choose j}\right]^{1/n}=4$

Question

Prove that
$$\lim_{n\to\infty}\Bigg[\sum_{i,j}{2i\choose i}{2j\choose j}\Bigg]^{1/n}=4$$ such that, $i+j=n$ and $0\le i,j \le n$.

My idea was to first sum the series when $i<j$ and when $j<i$. But I am unable to proceed. Please enlighten me.

Answer 1

Note that for any real $ \alpha $ : $$ \left(\forall x\in\left]-1,1\right[\right),\ \displaystyle\sum_{n=0}^{+\infty}{\displaystyle\binom{\alpha}{n}x^{n}}=\left(1+x\right)^{\alpha} $$ Thus, for $ x\in\left]-1,1\right[ $ : $$ \displaystyle\sum_{n=0}^{+\infty}{\left(-1\right)^{n}\displaystyle\binom{-\frac{1}{2}}{n}x^{n}}=\displaystyle\frac{1}{\sqrt{1-x}} \ \ \ \ \left(*\right) $$ Observe that : $$ \left(\forall n\in\mathbb{N}\right),\ \left(-1\right)^{n}\displaystyle\binom{-\frac{1}{2}}{n}=\displaystyle\frac{\left(-1\right)^{n}}{n!}\displaystyle\prod_{k=0}^{n-1}{\left(-\displaystyle\frac{1}{2}-k\right)}=\displaystyle\frac{1}{2^{n}n!}\displaystyle\prod_{k=0}^{n-1}{\left(2k+1\right)}=\displaystyle\frac{1}{4^{n}}\displaystyle\binom{2n}{n} $$ Multiplying the expression $ \left(*\right) $ by itself gives the following : $$ \displaystyle\sum_{n=0}^{+\infty}{\displaystyle\frac{1}{4^{n}}\left(\displaystyle\sum_{k=0}^{n}{\displaystyle\binom{2k}{k}\displaystyle\binom{2n-2k}{n-k}}\right)x^{n}}=\displaystyle\frac{1}{1-x}=\displaystyle\sum_{n=0}^{+\infty}{x^{n}} $$ Then we get that : $$ \left(\forall n\in\mathbb{N}\right),\ \displaystyle\sum_{k=0}^{n}{\displaystyle\binom{2k}{k}\displaystyle\binom{2n-2k}{n-k}}=4^{n} $$

Hence, $$ \left(\forall n\in\mathbb{N}^{*}\right),\ \left(\sum\limits_{i+j=n}{\binom{2i}{i}\binom{2j}{j}}\right)^{\frac{1}{n}}=4 $$

Answer 2

Since$$\sum_{ij}\binom{2i}{i}\binom{2j}{j}x^{i+j}=\left(\color{blue}{\sum_i\binom{2i}{i}x^i}\right)^2,$$the left-hand side's radius of convergence is the same as in the blue expression. Stirling's formula shows that, for large $i$,$$\binom{2i}{i}\sim\frac{4^i}{\sqrt{i\pi}}.$$Thus the radius of convergence is $\frac14$, which implies the desired result. In fact we can obtain the result without Stirling, since$$\frac{\binom{2i+2}{i+1}}{\binom{2i}{i}}=\frac{(2i+1)(2i+2)}{(i+1)^2}\sim4.$$

Answer 3

It is readily seen by the expansion of $(1+1)^{2i}$ that $$ \frac{{4^i }}{{2i + 1}} \le \binom{2i}{i} \le 4^i . $$ Hence, we have $$ \sum\limits_{i,j} \binom{2i}{i}\binom{2j}{j} = \sum\limits_{i = 0}^n \binom{2i}{i}\binom{2n - 2i}{n - i} \le \sum\limits_{i = 0}^n {4^i 4^{n - i} } = 4^n (n+1) $$ and $$ \sum\limits_{i,j} \binom{2i}{i}\binom{2j}{j} = \sum\limits_{i = 0}^n \binom{2i}{i}\binom{2n - 2i}{n - i} \\ \ge \sum\limits_{i = 0}^n {\frac{{4^i 4^{n - i} }}{{(2i + 1)(2n - 2i + 1)}}} \ge 4^n \sum\limits_{i = 0}^n {\frac{1}{{(n + 1)^2 }}} = \frac{{4^n}}{{n + 1}}. $$ It is now easy to see that the required limit is indeed $4$.

Answer 4

$$\begin{aligned}\lim_{n\to \infty}\left[\sum_{i+j=n\\ 0\le i,j\le n}{2i\choose i}{2j\choose j}\right]^{1/n}&=\left[\left(x^{i} \text{ in }(1+x)^{2i}\right)\left(x^{j} \text{ in }(1+x)^{2j}\right)\right]^{1/n}\\&= \left(x^{i+j} \text{ in }(1+x)^{2(i+j)}\right)^{1/n}\\ &= \left(x^{n}\text{ in } (1+x)^{2n}\right)^{1/n}=\lim_{n\to \infty}\left[{2n\choose n}\right]^{1/n}\end{aligned}$$

Now it is a trivial task to show that $\left[{2n\choose n}\right]^{1/n}\to 4$ as $n\to \infty$.

Edit: I must add that the trivial task of the computation of the limit of the binomial coefficient, can be done by a simple application of Stirling's Approximation.