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

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]

• @Domen Thanks for the edit. It is really useful
– Max
Jul 28, 2023 at 16:57
• Not $3n$ but $3\ n$. Jul 28, 2023 at 16:59
• @Max, do you have any reason to assume/believe that this should converge to a finite number? Glancing at the growth rate (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 $\infty$. Jul 28, 2023 at 17:04
• @Domen Can you please show how the limit is $\infty$?
– Max
Jul 28, 2023 at 17:11
• A search of OEIS identifies the sequence as A005259 "Apery numbers". The asymptotic expansion is given as (1+Sqrt[2])^(4*n+2)/(2^(9/4)*Pi^(3/2)*n^(3/2)). Jul 28, 2023 at 18:53

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]