Let $\ c\in \mathbb R.\ $ Let


$$ D\_n(c)\ := \frac {4^n}{\binom {2\cdot n}n\cdot\sqrt{4\cdot n + c}} $$


Then, more or less by the Wallis product theorem, we have this well-known convergence:


$$ \lim\_{n\rightarrow\infty}\ D\_n(c)\ \ =\ \ \frac {\sqrt{\pi}}2 $$


This holds regardless of the choice of the constant $\ c,\ $ however the quality of this convergence does depend on $\ c.$


 



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**QUESTION**   What value of $\ c\ $ leads to the fastest convergence?



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I'd conjecture that $\ c:=1\ $ (I may even run some numerical experiments). This is due to my method from my other ***Question*** about the accelerating the Wallis product. On the other hand, by passing from the factors to their exponents (Euler's style) one may end up with something like $\ e^\gamma\ $ somewhere inside that constant $\ c,\ $ who knows?



Approximating the central binomial coefficient
We have $$\left(\frac{D\_{n+1}(c)}{D\_n(c)}\right)^2=\frac{4(4n+c)(n+1)^2}{(2n+1)^2(4n+c+4)}=1+\frac{4n(c-1)+3c-4}{(2n+1)^2(4n+c+4)},$$
the convergence is the fastest if this is most close to 1, i.e. for $c=1$. In this case we get $$\frac{D\_{n+1}(1)}{D\_n(1)}=1+O(n^{-3}),\\ \frac{2D\_n(1)}{\sqrt{\pi}}=\prod\_{k=n}^\infty \frac{D\_k(1)}{D\_{k+1}(1)}=\exp\left(\sum\_{k\geq n} O(k^{-3})\right)=1+O(n^{-2}).$$


