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$$\sum_{n=0}^{\infty} 2^n = -1$$

I saw it on a 3blue1brown video and would like to learn more about it. Thank you.

Jam
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Nichole Kight
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    It might be noteworthy to note this comes up in the context of $p$-adic analysis in the video (if I remember correctly). As presented it suggests we're operating in the ordinary real numbers, where the summation is obviously false. I'm also not sure if it's something we could properly explain in a single post here... – PrincessEev Apr 18 '20 at 21:03
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    I would guess it's a reference to extending $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$ beyond $|x| < 1$ – Mason Apr 18 '20 at 21:04
  • The video in question. – Arthur Apr 18 '20 at 21:05
  • If you pretend that the sum $1+2+4+8+\dots$ exists, and you double it by doubling every term, you get $2+4+8+\dots$, which is the same sum except that the first term $1$ is missing. So the sum $S$ should satisfy $2S=S-1$. And the solution of that equation is $S=-1$. – Andreas Blass Apr 18 '20 at 22:20
  • Duplicate question: https://math.stackexchange.com/questions/37327/infty-1-paradox – Jam Oct 22 '22 at 18:47

1 Answers1

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The equality as stated makes no sense. The sum is divergent and doesn't have a finite value.

That being said, taking the limit of partial sums is only one of many ways to take a series and assign to it a real number that "represents" that series. And there are many ways to do this that happens to agree with the standard limit of partial sums on any series where they both are defined. These are often called "sums" for this reason, and they also often play nicely with intuitive algebraic manipulations.

Like using the standard trick for finding the sum of a convergent geometric series and applying it to a divergent series. That is one way get $-1$ from "adding" all powers of $2$. It is not the sum of all positive powers of $2$, but it is a real number that is naturally associated to the series of positive powers of $2$.

It might also be an idea to point out that the video in question ends with a short introduction to the $2$-adic integers, which is a construction where larger and larger powers of $2$ are considered "smaller", giving rise to a precise context where $\sum_{n=0}^\infty 2^n$ actually converges, and it indeed converges to $-1$. It also allows infinite digits extending to the left, rather than conventional numbers which allow infinite digits expansion to the right, although you really should use binary numbers if you go down that road.

Arthur
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    or a notional infinite-bit binary number written using 2's complement representation perhaps? https://en.wikipedia.org/wiki/Two%27s_complement – James Arathoon Apr 18 '20 at 22:10
  • @JamesArathoon $k$-bit 2's complement is just a fancy way of saying "modulo $2^k$" (some of my exams would've been much easier if I had been told that half a decade earlier than I was), and taking modulo "infinite power of $2$" is most conventionally done in a way called the $2$-adic numbers. I added a paragraph about it. – Arthur Apr 18 '20 at 22:32