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So I woke up this morning and I was thinking about the infinite product $-1 \times -1 \times -1 \times\dots$, and what it equals. I came to the conclusion that it equals $-i$. Alternatively stated,

$ {\displaystyle \prod_{i}^{\infty} (-1)} = -i $

Here's how I reached this:

$${\prod_{i}^{\infty} (-1)} = e^{\ln({\displaystyle \prod_{i}^{\infty} (-1)})} = e^{\displaystyle \sum_{i}^{\infty}{\ln(-1)}}= e^{\displaystyle \sum_{i}^{\infty}{i\pi}}=e^{i\pi\displaystyle \sum_{i}^{\infty}{1}}$$

Now, here's where I'm a little hesitant. I want to say that, from $\zeta(0)=-\frac{1}{2}$, we can conclude that

$$e^{i\pi\displaystyle \sum_{i}^{\infty}{1}} = e^{-\frac{1}{2}i\pi} = -i$$.

I have been told before that the sum $\displaystyle \sum_{i}^{\infty}{1}$ is not actually $-\frac{1}{2}$, but I'm not really sure why. It would seem that if this is the case, then my product would in fact not be $-i$. Though, I must say that $-i$ sort of makes sense, because multiplying complex numbers is essentially rotating them, and so rotating by $180$ every time will get you $180+180+180+...$ is the same as $180*(1+1+1+...)$ which is (if my premise is right) $180*(-\frac{1}{2})=-90$. $-90$ degrees on the complex plane turns out to be $-i$.

So my question is, is there a hole in my logic? I know what not accounting for $\zeta(0)=-\frac{1}{2}$, the sum $1+1+1+...$ is divergent, but taking that into account, can I say with confidence that $-1 \times -1 \times -1 \times\dots = -i$?

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    $\begingroup$ What you've actually found is the value of $e^{i \pi \zeta(0)}$, where $\zeta$ is the Riemann Zeta Function $\endgroup$ Apr 25, 2018 at 17:59
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    $\begingroup$ Infinite products are (usually) defined as the limit of the sequence of partial products. There this infinite product is not convergent. In fact it agrees with some of your observations since the parity of $\sum_{k=1}^n1$ alternate between odd and even as n changes. $\endgroup$
    – Bumblebee
    Apr 25, 2018 at 18:12
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    $\begingroup$ In you interpret $\prod_{n=1}^\infty (-1)$ as the ratio of two zeta-regularized product, $\prod_{n=1}^\infty (e^{i\pi} n)$ and $\prod_{n=1}^\infty n$, the value of your product do equal to $e^{i\pi\zeta(0)} = -i$. $\endgroup$ Apr 25, 2018 at 18:18
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    $\begingroup$ You should not use $i$ as the index in the product and the number such that $i^2+1=0$. $\endgroup$ Oct 12, 2020 at 14:12

1 Answer 1

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$$\prod_{n=1}^{\infty}(1+c)=\sum_{n=0}^{\infty}(2n)!/(n!)^2*(-c/4)^n=\frac{1}{\sqrt{(1+c)}}$$

The problem i've is that I found $-1=e^{i\pi}$ why wouldn't it be $-1=e^{3i\pi}$?

And thuse according to your logic the answer can also be i.

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  • $\begingroup$ You start with a formal product $\prod (1+c)$ then what formal transformation are you doing. $\endgroup$
    – reuns
    May 25, 2020 at 20:31
  • $\begingroup$ I assumed the question was the regularised product because else it's obvious divergent. But i applied partitions/writing a product as sums, sums you regularise to zeta(0) in this case. But the argument still stands i think. The inbetween result was just an illustration. It needs more effort the root of -1 is either i or -i, so this question needs a better story as answer. $\endgroup$
    – Gerben
    May 26, 2020 at 1:37
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    $\begingroup$ So this is an answer to the question: "So my question is, is there a hole in my logic? " I think this is a hole in this logic. $\endgroup$
    – Gerben
    May 26, 2020 at 1:41

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