# Showing that infinite product $\prod{(1+\frac{i}{k})}$ diverges

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?

• You may want to clarify the index or indices. For example, are $i$ and $k$ indices? If so, do they both range over infinity? i.e. $$\prod \left(1+\frac{i}{k}\right)=\prod_{i,k\ge 0}\left(1+\frac{i}{k}\right)?$$ – 000 Dec 5 '12 at 3:06
• $i$ is the imaginary number. $k$ is the index. I was not sure how to add indices, but I had hoped it was clear from the question. I can see how that's confusing. – tacos_tacos_tacos Dec 5 '12 at 3:07
• Thanks! I edited your post for you to clarify this. :-) – 000 Dec 5 '12 at 3:09
• Are you certain this diverges? I may have an error in my proof, but my work indicates this is not the case. – 000 Dec 5 '12 at 3:19
• I am not certain that it diverges, but Bak and Newman are... – tacos_tacos_tacos Dec 5 '12 at 3:21

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.