# Is the infinite product of $-1 \times -1 \times -1 \times\dots = -i$?

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,

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

Here's how I reached this:

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

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 $-1 \times -1 \times -1 \times\dots = -i$?

• What you've actually found is the value of $e^{i \pi \zeta(0)}$, where $\zeta$ is the Riemann Zeta Function Apr 25, 2018 at 17:59
• 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. Apr 25, 2018 at 18:12
• 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 $e^{i\pi\zeta(0)} = -i$. Apr 25, 2018 at 18:18
• You should not use $i$ as the index in the product and the number such that $i^2+1=0$. Oct 12, 2020 at 14:12

$$\prod_{n=1}^{\infty}(1+c)=\sum_{n=0}^{\infty}(2n)!/(n!)^2*(-c/4)^n=\frac{1}{\sqrt{(1+c)}}$$
The problem i've is that I found $$-1=e^{i\pi}$$ why wouldn't it be $$-1=e^{3i\pi}$$?
• You start with a formal product $\prod (1+c)$ then what formal transformation are you doing. May 25, 2020 at 20:31