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

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?

share|cite|improve this question
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
up vote 6 down vote accepted

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.

share|cite|improve this answer

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.