$\lim_{n\to\infty}\left(\sum_{0\lt{i},j\lt{n}}{\binom{2i}{i}\binom{2j}{j}}\right)^\frac{1}{n}$

Question

Find the value of the limit $$\lim_{n\to\infty}\left(\sum_{0\lt{i},j\lt{n}}{\binom{2i}{i}\binom{2j}{j}}\right)^\frac{1}{n}$$ Attempt: This changes to $$\lim_{n\to\infty}\left(\sum_{0\lt{i}\lt{n}}{\binom{2i}{i}}\right)^\frac{2}{n}$$ Now taking log it becomes $$\lim_{n\to\infty}\frac{2}{n}\log\left(\sum_{0\lt{i}\lt{n}}{\binom{2i}{i}}\right)$$ Now I am not able observe some adjustment and converting it into a definite integral which is usually the most reliable way. I am also having an intuition that it can be related by combinatorics to find a closed form of the sum. Please help to proceed.

Thanks

Answer 1

Because the sum is inside a logarithm, we only need very loose bounds which follow from the fact that $\binom{2n-2}{n-1}$ is the largest of the $2n-1$ positive numbers $\binom{2n-2}i$ ($0\le i\le 2n-2$) that sum to $2^{2n-2}$: \begin{align*} \sum_{i=1}^{n-1}\binom{2i}{i} &\ge \binom{2n-2}{n-1} \ge \frac1{2n-1} \cdot 2^{2n-2} \\ \sum_{i=1}^{n-1}\binom{2i}{i} &\le (n-1)\binom{2n-2}{n-1} \le n\cdot 2^{2n-2}. \end{align*} Therefore \begin{align*} \lim_{n\to\infty} \frac2n \log \frac{2^{2n-2}}{2n-1} &< \lim_{n\to\infty} \frac2n \log \sum_{i=1}^{n-1} \binom{2i}{i} < \lim_{n\to\infty} \frac2n \log (n\cdot 2^{2n-2}) \\ \lim_{n\to\infty} \frac2n \bigl( (2n-1)\log 2 - \log(2n-1) \bigr) &< \lim_{n\to\infty} \frac2n \log \sum_{i=1}^{n-1} \binom{2i}{i} < \lim_{n\to\infty} \frac2n \bigl( (2n-2)\log 2+\log n \bigr) \\ \end{align*} and it's straightforward from here to verify that the limits on both sides (and therefore the limit in the middle) equal $4\log 2$. We conclude that $$ \lim_{n\to\infty}\left(\sum_{0\lt{i},j\lt{n}}{\binom{2i}{i}\binom{2j}{j}}\right)^\frac{1}{n} = e^{4\log2} = 16, $$ which is confirmed by numerical calculation.

Answer 2

Since $$\binom{2 i}{i}=\frac{2^{2 i} \,\Gamma \left(i+\frac{1}{2}\right)}{\sqrt{\pi } \,\, \Gamma (i+1)}$$ the summation over $i$ would lead to a nasty Gaussian hypergeometric function.

Have a look at sequence $A006134$ in $OEIS$ and in particular at the second plot which shows that $$\log\left(\sum_{i=0}^n \binom{2 i}{i}\right)$$ is almost a straight line going through origin.

If the slope is $\alpha$, then $$\frac 1{2n}\log\left(\sum_{i=0}^n \binom{2 i}{i}\right)=\frac \alpha 2$$ As we can expect $\alpha$ is slightly smaller that $2\log(2)$. Then, the limit.

Computing $$S_n=\frac 1{2n}\log\left(\sum_{i=0}^n \binom{2 i}{i}\right)$$ for $n=10^k$ $$\left( \begin{array}{cc} k & S_{10^k} \\ 0 & 0.549306 \\ 1 & 0.621651 \\ 2 & 0.680213 \\ 3 & 0.691278 \\ 4 & 0.692903 \\ 5 & 0.693117 \\ \end{array} \right)$$