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]
  • $\begingroup$ @Domen Thanks for the edit. It is really useful $\endgroup$
    – Max
    Jul 28 at 16:57
  • $\begingroup$ Not $3n$ but $3\ n$. $\endgroup$ Jul 28 at 16:59
  • 1
    $\begingroup$ @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 $\infty$. $\endgroup$
    – Domen
    Jul 28 at 17:04
  • $\begingroup$ @Domen Can you please show how the limit is $\infty$? $\endgroup$
    – Max
    Jul 28 at 17:11
  • 3
    $\begingroup$ 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)). $\endgroup$
    – Somos
    Jul 28 at 18:53

1 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]


  • $\begingroup$ +1 and accepted $\endgroup$
    – Max
    Jul 29 at 6:32

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