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Can someone explain why this occurs. I came across this in a book by Titchmarsh.

$$\prod_{n=2}^{\infty}\left(1-\frac{e^{in\theta}}{\log(n)}\right)$$

this sum does not converge for any rational value of $\theta/\pi$ but is convergent if $\theta/\pi$ is a certain algebraic number

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  • $\begingroup$ Hint: in the case it's rational, $e^{in\theta}$ will be $\pm 1$ infinitely often. Take logs of the product and look at a second order Taylor expansion which has terms $1/\log^2(n)$: the sum should diverge. $\endgroup$ – Alex R. Mar 10 '14 at 3:45
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One should look up "on a class of conditionally convergent infinite products" it is a paper by Littlewood, this contains a neat answer. He proves that for any product of the form $$\prod_{n=1}^{\infty}\left(1+a_ne^{i n\theta}\right)$$ is convergent for certain algebraic values of $\theta/\pi$ with the condition that then $a_n$ are positive and strictly decreasing. Note, the motivation for this was another interesting question that was posed by Hardy.

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