TITLE: Showing that infinite product $\prod{(1+\frac{i}{k})}$ diverges
QUESTION [7 upvotes]: In Bak and Newman's Complex Analysis they ask to show that the infinite product $\prod_{k \ge 1}{(1+\frac{i}{k})}$ diverges (with $i$ being the imaginary unit). My intuition is that it does not diverge to $0$, but rather just kind of oscillates randomly around the origin for large partial products. However, I am having a hard time proving this. If I break it down into two products of $r$ and $e^{i\theta}$ this doesn't help, because $\theta \rightarrow 0$ pretty clearly, and then I do not get my desired result of perpetual rotation. The $r$ term, $\prod_{k \ge 1}{\sqrt{1+\frac{1}{k^2}}}$ is not very informative either. I guess I have two questions: is my assumption that it oscillates kind of randomly at $\infty$ incorrect? If it is correct, how might I go about showing that this is the behavior?

REPLY [7 votes]: The product diverges but its norm converges.  So indeed it keeps "circling around".  To see that its norm converges observe that $$ 1 \leq \left| 1 + \frac{i}{k} \right| = \sqrt{1 +\frac{1}{k^2}} \leq 1 + \frac{1}{2k^2}$$ and the product $$ \prod_{k=1}^{\infty} \left(1 + \frac{1}{2k^2} \right) $$ converges.  However, for the argumen of $1 + \tfrac{i}{k}$ we have
$$
\tan \arg \left(1 + \frac{i}{k}\right) = \frac{1}{k}
$$
and therefore
$$
\arg \left(1 + \frac{i}{k}\right) \geq \frac{\pi}{4k}
$$
for all $k \geq 1$.  The argument of a partial product is
$$
\arg \prod_{k=1}^N \left( 1 +\frac{i}{k} \right) = \sum_{k=1}^N \arg \left( 1 +\frac{i}{k} \right) \geq \frac{\pi}{4} \sum_{k=1}^N \frac{1}{k}
$$
and the latter sum diverges.