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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 $\ c.$

 

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

 

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?

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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 $$\frac{D_{n+1}(1)}{D_n(1)}=1+O(n^{-3}),\\ \frac{2D_n(1)}{\sqrt{\pi}}=\prod_{k=n}^\infty \frac{D_k(1)}{D_{k+1}(1)}=\exp\left(\sum_{k\geq n} O(k^{-3})\right)=1+O(n^{-2}).$$

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  • $\begingroup$ Fedor, wonderful! $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 17, 2016 at 8:09
  • $\begingroup$ Thus now, in the textbooks, when they write about $\ \binom{2⋅n}n,\ $ they should replace $\ 2\cdot \sqrt{n}\ $ by $\ \sqrt{4\cdot n+1}$. $\endgroup$
    – Włodzimierz Holsztyński
    Commented Dec 17, 2016 at 8:14
  • $\begingroup$ @Fedor: This is neat. But, I have a question: don't you think you flipped the $1+O(1/n^3)$ in the infinite product. Plus, intuitively, in $D_n(c)$ the convergence is faster as $c$ grows instead of for small $c$. What do you think? $\endgroup$
    – T. Amdeberhan
    Commented Dec 17, 2016 at 8:32
  • $\begingroup$ @T.Amdeberhan I do not understand your question. What we do with O is, after all, only multiplying inequalities between positive quantities. $\endgroup$
    – Fedor Petrov
    Commented Dec 17, 2016 at 8:56
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    $\begingroup$ no, $(1+O(1/n^3))^{1/2}=1+O(1/n^3)$ $\endgroup$
    – Fedor Petrov
    Commented Dec 17, 2016 at 9:16

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