Calculate $\lim_{n\to\infty} \left(\frac{e}{3}\right)^{3n} a_n^4$

Question

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]

Answer 1

Let us leave only the term with r==n in the sum. Then

r = n; Limit[(Binomial[n, r] Binomial[n + r, r])^8*(E/3)^(3 n),  n -> Infinity]

\[Infinity]