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How can i calculate $\prod_{n=1}^{\infty}{erf(n)} $ with $erf(z) = \frac{2}{\sqrt{\pi}}\int_0^z e^{-z^{2}} \mathrm{d}z$? I know it's something like 0,84. And i see that only the first terms are important because $\lim\limits_{n \rightarrow \infty}{erf(n)}=1$

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By 'limes', did you mean '$\lim$s'? I suppose you don't just want to calculate it, but calculate it efficiently? –  David Roberts Nov 5 '12 at 21:45
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What you want is not clear: a closed form, an approximate value? Moreover, your remark on the first term is a mistake: that the limit is 1 is a necessary condition for the product to converge, but that does not mean that only the first terms matter. –  Benoît Kloeckner Nov 5 '12 at 21:56
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Convergence is very rapid. $0.83874032043948921151864887207058844734595614546070$ The ISC does not recognize it. isc.carma.newcastle.edu.au –  Gerald Edgar Nov 5 '12 at 22:26
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Note: "limes" is the Latin term, retained in some languages, but called "limit" in English. –  Gerald Edgar Nov 5 '12 at 22:28
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Ah, thanks, Gerald, that's interesting. I thought it perhaps just an mistake in plural formation. –  David Roberts Nov 6 '12 at 0:25

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