# convergence of an infinite product $\prod_{j=1}^{n}{(0.5+ \frac{1}{\pi}\arctan(jx))}$

Does the following infinite product converge as $$n \rightarrow \infty$$ : $$\prod_{j=1}^{n}{(0.5+ \frac{1}{\pi}{\arctan(jx)})}$$

If yes, then what is the limiting value?

I encountered the above while dealing with the Distribution Function of the Cauchy Random Variable.

Rewrite as Song did to the following form $$\prod_{n=1}^\infty \left(\frac12+\frac1\pi\arctan(nx)\right)=\prod_{n=1}^\infty\left(1-\frac1\pi\arctan\left(\frac1{nx}\right)\right).$$ Note that if $$x>0$$, then $$\arctan(1/(nx))>0$$ and (in particular, since $$\arctan(x)\approx x$$ around $$0$$. Hence, since all terms are now between $$0$$ and $$1$$, we invoke the monontone convergence theorem to find that this infinite product is convergent.
For negative $$x$$, we invoke the following lemma (which you can easily prove):
Lemma: For non-negative sequences $$(a_i)_{i=1}^{\infty}$$, we have $$\prod_{i=1}^{\infty}(1+a_i) < \infty$$ if and only if $$\sum_{i=1}^{\infty} a_i < \infty.$$
Thus, we need to find whether or not $$\sum_{n=1}^{\infty} \arctan\left(\frac{1}{n x}\right) = \sum_{n=1}^{\infty} \frac{1}{nx}\cdot \left(n x\arctan\left(\frac{1}{n x}\right)\right)$$ is convergent. Since $$nx\arctan(1/(nx))$$ converges to $$1$$, this series behaves as $$\sum_{n=1}^{\infty} 1/(nx)$$, which is definitely not convergent.