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Does the infinite product diverge to zero or some other value?

$$\prod_{n \mathop = 1}^\infty {\frac{2^n}{3^n}}$$

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    $\begingroup$ It's an infinite product, not a series. $\endgroup$ – Robert Israel Jul 19 '13 at 4:49
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    $\begingroup$ The correct terminology is : "it diverges to 0", consider taking the logarithm of this product and looking at the sum. $\endgroup$ – Arjang Jul 19 '13 at 5:26
  • $\begingroup$ Thank you for correcting my amateurish mistakes. $\endgroup$ – KeithSmith Jul 19 '13 at 12:10
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Consider $$P_m=\prod_{n=1}^m\frac{2^n}{3^n}=\left(\frac{2}{3}\right)^{m(m+1)/2}$$

Then, $$P_{\infty}=\prod_{n=1}^{\infty}\frac{2^n}{3^n}=\lim_{m\to\infty}P_m=\lim_{m\to\infty}\left(\frac{2}{3}\right)^{m(m+1)/2}=0$$

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    $\begingroup$ Why is it so big? $\endgroup$ – Potato Jul 19 '13 at 5:21
  • $\begingroup$ @Potato: with \frac it's too small and with \dfrac it's too big. $\endgroup$ – Aang Jul 19 '13 at 5:24
  • $\begingroup$ I personally don't think it's too small with \frac. But to each their own, I suppose. $\endgroup$ – Potato Jul 19 '13 at 5:26
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    $\begingroup$ You can scale rendered math for yourself if you find it too small. There's no need to use commands like \large, really. $\endgroup$ – Antonio Vargas Jul 19 '13 at 7:00
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    $\begingroup$ I guess, @martycohen, that’s Littlewood. $\endgroup$ – Lubin Jul 24 '13 at 4:41

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