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?
Showing that infinite product $\prod{(1+\frac{i}{k})}$ diverges
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complex-analysis
elementary-number-theory
infinite-product
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0You 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. $$\prod \left(1+\frac{i}{k}\right)=\prod_{i,k\ge 0}\left(1+\frac{i}{k}\right)?$$ – 2012-12-05
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0$i$ 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. – 2012-12-05
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0Thanks! I edited your post for you to clarify this. :-) – 2012-12-05
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0Are you certain this diverges? I may have an error in my proof, but my work indicates this is not the case. – 2012-12-05
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1I am not certain that it diverges, but Bak and Newman are... – 2012-12-05
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0Also, if you just evaluate this product in a calculator, it does not seem to converge, but there is always the possibility I guess that it takes a really long time to converge – 2012-12-05
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0Welp. I gave it my third try. Critique away! :-) – 2012-12-05
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0@Limitless, even though your third try was incorrect, the critical idea was there, which was that the argument of the product was given by the sum of the arguments of the components – 2012-12-05
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0Thanks for the vote of confidence. :-) – 2012-12-05
1 Answers
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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.