Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $G=0$.

share|cite|improve this question
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 $(v_n)$.

share|cite|improve this answer

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.