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Edited: Making the question as brief as possible to avoid future confusion and misunderstanding.

Note

This was moved as a separate question from: Product of all real numbers in a given interval $[n,m]$

Since it was a part of it that wasn't getting any attention.

Question

How would one calculate the infinite product of negative numbers? For example, in this case:

$$-1\times -1\times-1\times -1\dots=$$

Or is the result of this series simply undefined?

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Vepir
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  • Saying "... it has 2 possible outcomes if the infinite sequence ends with $-1$ or $1$" is a fallcy. – Edward Evans Apr 17 '16 at 15:59
  • we can write it as $i^{2n}$ so can be $\pm 1$ – Archis Welankar Apr 17 '16 at 16:13
  • @ArchisWelankar More like $(-1)^{\infty}=i^{2\times\infty}$ ? – Vepir Apr 17 '16 at 16:24
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    $\lim\limits_{n\to\infty}(-1)^n$ is undefined. – barak manos Apr 17 '16 at 16:25
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    no there isnt such thing its an indeterminate form so can be $\pm 1$ implying it cant be found out – Archis Welankar Apr 17 '16 at 16:25
  • @barakmanos, I would satisfy myself with that as an answer if you post it as one, unless someone somewhere actually used something to "define" it like $0^0$ is sometimes defined $1$. – Vepir Apr 17 '16 at 16:44
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    I know that Numberphile are just trying to excite people about math and contribute to the general excitement about math as a cool subject. But I find their methods to end up as more confusing than enlightening to people who are less familiar with mathematics (read: they approach it as the physicists I believe they are). The fact you both claim that you're looking for something akin to Cesaro summation, and have accepted an answer stating that $\lim (-1)^n$ does not exist (just like the limit of $(1,0,1,0,1,\ldots)$ does not exist) is a testament to this added confusion. – Asaf Karagila Apr 17 '16 at 17:04
  • $$\Pi_{n=1}^\infty(-1)=e^{\ln(\Pi_{n=1}^\infty(-1))}=e^{\sum_{n=1}^\infty\ln(-1)}$$ $$=e^{\ln(-1)\sum_{n=1}^\infty1}$$ $$=e^{\ln(-1)(1/2)}=(-1)^{1/2}=\pm i?$$ – Simply Beautiful Art Apr 26 '16 at 22:07

2 Answers2

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How would one calculate, or attach a value to the infinite product of negative numbers?

For example, in this case: $(-1)\times(-1)\times(-1)\times(-1)\ldots=$?

The value of $\lim\limits_{n\to\infty}(-1)^n$ is undefined.

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barak manos
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You could attach some the value of a limit of the average of the values if such a limit of averages would converge. For example arithmetic average of +1-1+1-1... would be 0 as when dividing by the number n of terms so far the oscillations would get below any $\epsilon\in \mathbb{R}$.

mathreadler
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  • Your question is very loosely worded and unspecified "how could one attach a value for this". Maybe you should specify it a bit more if you want a more specific answer. – mathreadler Apr 17 '16 at 16:24
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    @Matta You're being quite rude to a person that actually tried to help you. – Eff Apr 17 '16 at 16:27
  • @Matta: "Who said anything about needing an arithmetic average of anything?" - you did: "Could we then say that this product is 0? By taking the average of −1 and 1 in our efforts to assign a value to it?". – barak manos Apr 17 '16 at 16:29
  • You literally wrote "How would one calculate, or attach a value to...". Most people would probably say you can't, as it doesn't converge. If you want to assign a value in some other sense maybe you should bother explaining what properties you want that assignment to have. – mathreadler Apr 17 '16 at 16:29
  • @Matta: You don't need to explain the term average. You asked (very rudely if I may add) "Who said anything about average" while you did so yourself. If you have any refinement to your original description, then please apply it in your question, and don't take it out on the person who's trying to help you with it! – barak manos Apr 17 '16 at 16:38
  • @barakmanos I'm sorry for not picking the right words since I reacted a bit too fast. English is not my first language so posting this at first didn't seemed rude or to conflict with the phrases I used in my question. Again I'm sorry for the big misunderstandment. – Vepir Apr 17 '16 at 16:40
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    @Matta: No problem :) Please note that $\lim\limits_{n\to\infty}(-1)^n$ is simply undefined. – barak manos Apr 17 '16 at 16:40
  • There are very many types of averages, I just gave one example. It was intended to help you get new thoughts and ideas. – mathreadler Apr 17 '16 at 16:45
  • Cesàro summation is a version of what I proposed, a weighted sum. Yes I agree I could have written a comment with the content that I had written so far. Sometimes people on this site write very short answers which they improve over time. – mathreadler Apr 17 '16 at 17:04