I came across this identity: $$\sum_{k=0}^n \frac{n!(n+1)!}{(n-k)!(n+k+1)!}(2k+1) = n+1. $$ This shows up quite undirectly when considering spherical harmonics, I would be interested in a direct argument.
More importantly, suppose $$A_n = \sum_{k=0}^n \Big(\frac{n!(n+1)!}{(n-k)!(n+k+1)!}\Big)^2(2k+1) \qquad \text{and} \qquad B_n = \sum_{k=0}^n \Big(\frac{n!(n+1)!}{(n-k)!(n+k+1)!}\Big)^2(2k+1)^2.$$ Do we have $\frac{B_n}{A_n}\to \infty$ ?
The Cauchy-Schwarz inequality implies that $B_n\ge n+1$, using the above identity. I am not sure whether $A_n$ diverges or not. The factorial quotient gets very small as $k$ approaches $n$.
Edit: I have accepted Claude's answer for the idea of using hypergeometric functions and their values, however I still welcome references for more details about these values, or alternative proofs.
I think that, if you compute the first $A_n$ and $B_n$ you should see that they are not of the same degree of magnitude at all.
Let
$$ a_n=\Bigg(\frac{n!(n+1)!}{(n-k)!(n+k+1)!}\Bigg)^2(2k+1)$$ $$ b_n=\Bigg(\frac{n!(n+1)!}{(n-k)!(n+k+1)!}\Bigg)^2(2k+1)^2$$ and consider $$A_n(x)=\sum_{k=0}^n a_n\, x^n \qquad \text{and} \qquad B_n(x)=\sum_{k=0}^n b_n\, x^n$$ Both are given in terms of hypergeometric function.
$$A_n(1)= \frac{(n+1)^2}{2 n+1}$$ $$B_n(1)=\pi (n+1) \frac{ \Gamma (n+2)\,\, \Gamma \left(2 n+\frac{1}{2}\right)}{2^{2 n+1}\,\,\Gamma \left(n+\frac{1}{2}\right)^2 \,\,\Gamma \left(n+\frac{3}{2}\right)}$$
Asymptotically, using Stirling approximation and Taylor series, $$R_n=\frac{B_n(1)}{A_n(1)}=\sqrt{\frac{\pi n}{2}} \left(1+\frac{3}{16 n}+\frac{9}{512 n^2}+\frac{71}{8192 n^3}+O\left(\frac{1}{n^4}\right)\right)$$ which is very accurate.
For example, using $n=10$, the exact result is $$R_{10}=\frac{240990435}{59676188}=4.038301$$ while the approximation gives $$\frac{8347111 }{1638400}\sqrt{\frac{\pi }{5}}=\color{red}{4.0383}71$$