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I dont know how to evaluate the first one, for second one I can only show the sum is less than 2. $$\begin{align} & \prod\limits_{k=4}^{\infty }{\left( 1-{{\left( \frac{3}{k} \right)}^{3}} \right)} \\ & \sum\limits_{n=1}^{\infty }{\sin \left( \frac{1}{2^{n-1}} \right)} \\ \end{align}$$

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You shouldn't be able to show that because the sum diverges clearly via small-angle approximations. For the first one, just take the exponential of its logarithm. – Yonatan N Jan 30 '13 at 5:01
@YonatanN Sry, I got a typo! – gauss115 Jan 30 '13 at 5:05
up vote 1 down vote accepted

Here is a result by maple for the infinite product

$$\prod\limits_{k=4}^{\infty }{\left( 1-{{\left( \frac{3}{k} \right)}^{3}} \right)} = {\frac {8}{15561\,\pi}}\,{ {\cosh \left( \frac{3\pi \,\sqrt {3}}{2} \right) }}. $$

Added: The numerical sum of the series evaluated by maple

$$ \sum\limits_{n=1}^{\infty }{\sin \left( \frac{1}{2^{n-1}} \right)} \sim 1.817928721 $$

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Thx, but how to evaluate the first one? :) – Ryan Jan 30 '13 at 12:35

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