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On the Wolfram page about pi formulas, there is this curious limit by R. W. Gosper (130) $$\lim\limits_{n\to\infty}\prod\limits_{k=n}^{2n}\dfrac{\pi}{2\arctan k}=4^{1/\pi}.$$

The only reference given is an entry from 1996 in some forum. Has anybody a proof or reference for this or similar formulas?

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  • $\begingroup$ Are you asking for a proof of this limit? $\endgroup$
    – GH from MO
    Commented Mar 9, 2012 at 19:42
  • $\begingroup$ yes, this was intended. $\endgroup$
    – Wolfgang
    Commented Mar 9, 2012 at 20:43

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Here is a sketch, which can be completed to a proof.

Asymptotically (i.e. for $k$ large), $\ln\left(\frac{\pi}{2\arctan k}\right) = \frac{2}{\pi k} + O(\frac{1}{k^2})$. Sum termwise to get $$\frac{2\Psi(2n+1)}{\pi} - \frac{2\Psi(n)}{\pi} + \sum_{k=n}^{2n} O(\frac{1}{k^2})$$ Use the propreties of $\Psi$ to simplify this to $$\frac{2\ln 2}{\pi} + \frac{\Psi(n+1/2)}{\pi} -\frac{\Psi(n)}{\pi} + \sum_{k=n}^{2n} O(\frac{1}{k^2}) $$ Now, for $n$ large, this is asymptotically $$\frac{2\ln 2}{\pi} + O(\frac{1}{n}).$$

All the terms in the product at positive, so taking the logarithm was legitimate. The termwise sum can similarly be justified.

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  • $\begingroup$ Can you explain the notation $\Psi(x)$? $\endgroup$
    – GH from MO
    Commented Mar 9, 2012 at 21:29
  • $\begingroup$ $\Psi$ is the Digamma function - en.wikipedia.org/wiki/Digamma_function, the logarithmic derivative of $\Gamma$. $\endgroup$
    – Jacques Carette
    Commented Mar 9, 2012 at 22:03
  • $\begingroup$ Thanks! Your $\sim$ is confusing: $f\sim g$ means that $f/g\to 1$, while $f=g+O(h)$ means that $|f-g|/h$ is bounded, and $f\sim g+O(h)$ confuses the two. The $O$-notation is not about asymptotics but error terms (yielding asymptotics sometimes). Also, you can avoid using $\Psi$ as $\sum_{k=n}^{2n}\frac{1}{k}\to\int_{1}^2\frac{dx}{x}=\ln 2$ is immediate from the definition of the Riemann integral (Riemann sums approach the integral). $\endgroup$
    – GH from MO
    Commented Mar 10, 2012 at 0:18
  • $\begingroup$ Fixed the $\sim$, since that was needlessly confusing. Keeping the $\Psi$ as it is one of my favourite functions! $\endgroup$
    – Jacques Carette
    Commented Mar 10, 2012 at 1:47
  • $\begingroup$ All right. Also, your argument is not a sketch but a full proof. $\endgroup$
    – GH from MO
    Commented Mar 10, 2012 at 2:22

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