I need to calculate the limit $$\lim_{n\to\infty} \left(\frac{e}{3}\right)^{3n} a_n^4 $$ where $a_n=\sum_{r=0}^{n}\left(\binom{n}{r}\binom{n+r}{r}\right)^2$ and $e$ is the natural base of logarithm.
I managed to simplify $a_n$ with the following code:
Sum[(Binomial[n, r] Binomial[n + r, r])^2, {r, 0, n}]
$$a_n= {}_4F_3\left ( -n,-n,n+1,n+1;1,1,1;1 \right )$$ where ${}_4F_3$ represents the hypergeometric function.
So our limit becomes $$\lim_{n\to\infty} \left(\frac{e}{3}\right)^{3n}{}_4F_3^4\left ( -n,-n,n+1,n+1;1,1,1;1 \right ) $$
I tried a code in Mathematica but could not get an answer:
Limit[(E/3)^{3 n}*(HypergeometricPFQ[{-n, -n, n+1, n+1}, {1, 1, 1}, 1])^4,
n -> Infinity]
Table[{n, (E/3)^(3 n)*(HypergeometricPFQ[{-n, -n, n + 1, n + 1}, {1, 1, 1}, 1])^4}, {n, 0, 1000, 100}] // N // Grid), it feels like this limit is $\infty$. $\endgroup$(1+Sqrt[2])^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). $\endgroup$