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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