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Let $G=\prod_{n=1}^{\infty} \frac{2\tan^{-1}(v_{n})}{\pi}$, where $v_{n}$ is an increasing, monotonic sequence of natural numbers: is it true that there is no sequence of $v_{n}\in\mathbb{N}$ such that $G>1$ for sequences $v_{n}$ such that $G$ is nonzero? I know that $v_{n}$ has to be spread more thinly than just $v_{n}=n$ -- in that case $G=0$.

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up vote 6 down vote accepted

If you look at |G| it is product of factors of the form $|\frac{\arctan{x}}{\pi/2}|<1$, since arctan has its image the interval $(-\pi/2,\pi/2)$, and therefore $|G|<1$ for any sequence $(v_n)$.

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To ensure that $G\ne0$ is more interesting: this holds if and only if the series $\displaystyle\sum\frac1{v_n}$ converges.

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Perhaps needs some explanation. asymptotics of $\tan^{-1}(v)$ as $v \to \infty$. –  GEdgar Jun 19 '11 at 18:34
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