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.

Your last question is to compute the limit if it exists. I have to say, I don't know whether we could do this analytically.

  • $\begingroup$ (+1) Can you please direct me to a proof of the lemma . $\endgroup$ – John Feb 16 at 18:01
  • $\begingroup$ Try it yourself first ;) it is a neat little exercise. If you can't solve it, which direction of the if and only if are you troubling with? $\endgroup$ – Stan Tendijck Feb 16 at 22:11

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