I puzzled two high school Pre-calc math teachers today with a little proof (maybe not) I found a couple years ago that infinity is equal to -1:

  1. Let x equal the geometric series: $1 + 2 + 4 + 8 + 16 \ldots$

    $x = 1 + 2 + 4 + 8 + 16 \ldots$

  2. Multiply each side by 2:

    $2x = 2 + 4 + 8 + 16 + 32 \ldots$

  3. Again from the equation in step 1, move the $1$ term to the left hand of the equation:

    $x - 1 = 2 + 4 + 8 + 16 + 32 \ldots$

  4. So the following appears to be true:

    $2x = x - 1 \implies x = -1$

This is obviously illogical. The teachers told me the problem has to do with adding the two infinite geometric series, but they weren't positive. I'm currently in Pre-calc, so I have extremely little knowledge on calculus, but a little help with this paradox would be appreciated.

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    The geometric series you have is equivalent to evaluating $\frac1{1-x}$ at $x=2$; the $-1$ is valid for that formula, but you are using the series outside its region of validity, which is where the trouble lies. That should beg the question of why adding up a bunch of positives results in a negative... ;) – J. M. ain't a mathematician May 06 '11 at 03:57
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    In the real numbers, x does not exist. (The geometric series has no sum.) Perhaps you can use an extension of the real numbers that contains ∞ and makes some of those operations valid, in which ∞ *does* satisfy everything up to step 4, namely 2x=x-1. (2∞ = ∞ = ∞-1.) But such an extension would not have the property that a+b=a+c => b=c, so you still cannot "subtract both sides by x" and get step 5. – ShreevatsaR May 06 '11 at 04:04
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    Here is a simple explanation without using the radius of convergence concept for geometric series. The series you gave diverges to infinity, and you can't do arithmetic operation such as minus on infinity, asking $\infty - \infty$ equals is kinda invalid... – Shuhao Cao May 06 '11 at 04:09
  • A neat trick you can do with *Mathematica*: `SequenceLimit[Table[Sum[2^j, {j, 0, n}], {n, 0, 10}]]`. Unless you take into account what I said earlier, it doesn't make sense indeed. Now try replacing `2` in that snippet with `x`... – J. M. ain't a mathematician May 06 '11 at 04:19
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    A similarly wrong (but simpler) proof would go like this: $\infty+1$ is still $\infty$ since you can't make it any larger. But then we have $\infty = \infty+1$ and we subtract infinity from both sides proving $0=1$. – Fixee May 06 '11 at 06:09
  • For a more substantial reference that your teachers might enjoy, $p$-adic Numbers by Fernando Q. Gouvea, Problem 21 on page 20, which is to show that, in $\mathbf Q_p, \; \; 1 + p + p^2 + p^3 + \cdots = \frac{1}{1-p}.$ $$ $$ http://www.springer.com/mathematics/numbers/book/978-3-540-62911-5 – Will Jagy May 06 '11 at 06:50
  • [Here](http://math.stackexchange.com/questions/37180) is a related thread. – J. M. ain't a mathematician May 06 '11 at 12:46
  • Beware that $\infty-\infty$ is indeterminated. – JSCB Sep 13 '12 at 10:36
  • For example, $・・・0000\pmod {10^n}= 1+・・・9999\pmod {10^n}→ ・・・9999 =-1 \pmod {10^n} $. Since $・・・9999=-1$=max number=(infinity), therefore infinity $=-1$. – Takahiro Waki Jan 27 '17 at 09:22

6 Answers6


When we talk about an "infinite sum", we are really talking about a limit. In this case, we are talking about the limit of the "partial sums" of the series. The partial sums are: $$\begin{align*} s_1 &= 1;\\ s_2 &= 1+2;\\ s_3 &= 1+2+4;\\ &\vdots \end{align*}$$ That is, $s_n$ is the sum of the first $n$ summands in the series. When we talk about the "value" of a series (an infinite sum), we are really talking about the limit of the $s_n$: that is, a specific real number $L$ that the $s_n$ are approaching as $n\to\infty$. Or we say that a series "equals $\infty$" if the values of $s_n$ grow without limit.

When you say $x = 1+2+4+\cdots$, what you are really saying is that the limit of $s_n$. In this case, the limit of the $s_n$ does not exist, because $$\lim_{n\to\infty}s_n = \infty.$$ The values of $s_n$ get arbitrarily large as $n\to\infty$.

It is certanly true as well that the sum $2+4+8+\cdots$ is also $\infty$, since $2\times\infty = \infty$ (in the extended reals). And if you subtract one, then you still get $\infty$ because $\infty -1 = \infty$ (in the extended reals).

So you can write $2x = x-1$.

What you cannot do, however, is "subtract $x$ from both sides"; because that would be writing $$2\times\infty - \infty = \infty -1 -\infty$$ and the problem is that even in the extended reals, $\infty-\infty$ is undetermined. It does not equal anything, and certainly not zero. In short, you cannot just cancel infinities.

Ross Millikan
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Arturo Magidin
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    So it's ok to add two infinities together and get infinity, but not subtract two infinities? – Christian May 06 '11 at 04:22
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    @vorbis: There's a reason $\infty-\infty$ is termed an *indeterminate* form... ;) – J. M. ain't a mathematician May 06 '11 at 04:26
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    @vorbis5: Yes; the problem is that the addition of infinities is defined so that $\infty+a=\infty$ whenever $a$ is a real number of $\infty$. That means that there is nothing you can add to $\infty$ to get $0$ (which is the *real* meaning behind "subtracting"). – Arturo Magidin May 06 '11 at 04:27
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    @vorbis: If I have an infinite number of marbles, each either red or blue, I can give you all of them. Then ∞−∞ = 0. Or say I give you all of the blue marbles. Then ∞−∞ = ∞. Or say I give you all but 11 marbles. Then ∞−∞ = 11. So ∞−∞ could be *anything*. Ok... these symbols aren't right. – The Chaz 2.0 May 06 '11 at 04:38

Two comments. First, it is possible to get this sum to make sense if one changes the notion of limit. The $2$-adic numbers are equipped with a different absolute value than the usual absolute value on rational numbers, and using this absolute value the series actually does converge to $-1$.

Second, the series $1 + z + z^2 + ...$ defines what's called a holomorphic function on the complex numbers with absolute value less than $1$. On these numbers it can be identified with its sum $\frac{1}{1 - z}$, but the latter expression now makes sense for all values of $z$ not equal to $1$. This is a special case of what's known as analytic continuation of a holomorphic function.

More generally, the limit definition Arturo describes is not the only way to give meaning to an infinite sum. It is the standard way, and for almost all intents and purposes it is the most useful way, but it is not the only way. Alternate methods are covered at the Wikipedia article on divergent series, and lest you think this is just a curiosity, summing divergent series happens to be relevant to certain physical theories which give sums that would diverge if they weren't summed in a non-traditional way.

If you're not familiar with $2$-adic numbers, here's a very brief introduction and a very short "proof" of the above identity using them. Basically, they behave like the natural numbers (with zero), except that they have infinite binary expansions going to the left. The number $1 + 2 + 4 + 8 + ...$ has binary expansion


in the $2$-adic numbers, and then the proof that it is equal to $-1$ is just a matter of carrying infinitely many times (add $1$ to the above)!

Qiaochu Yuan
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  • In general, Is there a reason why the limit in an appropriate p-adic norm equals the analytic extension or is it that they just so happens to be equal to each other? –  May 06 '11 at 04:39
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    @Sivaram: yes. In both cases $L = \sum x^i$ needs to satisfy $(1 - x) L = 1$ (in the first case because the sequence converges and in the second case because of the identity principle). For algebraic functions such as $\sqrt{1 + x}$ it can happen that you get different rational integers for different $p$ (try $x = 15, p = 3, 5$). Of course you will still get something squaring to $1 + x$; in other words, you still get _some_ analytic continuation, but possibly different branches. – Qiaochu Yuan May 06 '11 at 04:46
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    +1 for the 2-adics. I'd almost forgotten about that one... Amusingly, $-1 = 1+2+4+8+\cdots$ is also what you get if you try using the binomial theorem on $(1-2)^{-1}$. – Arturo Magidin May 06 '11 at 04:58
  • Thanks. Can analytical extension of any function in general be thought of as convergence in some weird norm? –  May 06 '11 at 15:05
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    @Sivaram: that depends on what exactly you mean by "norm." If you mean the usual sense of absolute value on a field, then by Ostrowski's theorem there are no more weird ones on $\mathbb{Q}$: there's just the $p$-adic ones and the ordinary one. And in general limits with respect to these norms will land in different fields, so it's unclear how to compare them. – Qiaochu Yuan May 06 '11 at 15:11
  • Lol. "So let us do math in a different universe with p-adic numbers$\dots$" started the founders of divergent series calculation techniques. – Simply Beautiful Art Sep 10 '16 at 00:16
  • @Qiaochu Yuan - +1 for the answer. This is exactly the example I used in a popular article I wrote years ago ("Divergence is not the fault of the series," Pi Mu Epsilon Journal, 8 (fall 1988), 588-589). I noted the relation of the $n$-th partial sum to the representation of -1 in an $n$-bit machine using 2's complement notation for integer arithmetic (some extra heuristic since I was teaching some elementary computer science at the time). – Chris Leary Jul 17 '20 at 20:17

I happen to be sitting next to a group of math majors, and I decided to explain this 'cool new proof' of a 'previously unknown result' to them. Here's how that went:

$$" x = 1 + 2 + 4 + 8 + 16 + ... "$$ They are already suspicious - it is peculiar giving $\infty$ a name like $x$, and so I can tell they already know that something strange is afoot.

"Note that $2x = 2 + 4 + 8 + ... = x - 1$."

They don't seem to hate this idea yet. One asks me if we are interpreting a sequence of partial sums - a very natural question, as it's sort of the only way to make any sense of it. But then I say

"Note therefore that $x - 1 = 2x \rightarrow x = -1$.

Now they all stopped me - it's utter nonsense, they say. As much as I might like to try to recover the situation and pull the hood over their eyes, they have seen through it. Unfortunately, so have I. It simply doesn't make sense to subtract 'infinity' from 'infinity.'

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    Haha thats an awesome answer. Thank you, math majors :D – Christian May 06 '11 at 04:16
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    woah woah woah... why am I not there?!??! – Nicolas Villanueva May 06 '11 at 04:28
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    @Nicolas: If you were in Skiles, then you could have been there. In 12 hours - I will wash my hands of this school and prepare for the great beyond. – davidlowryduda May 06 '11 at 04:36
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    Truer words never spoken sir! So turns out that my mom's is coming into town tonight and sleeping in my room at North Ave, and I'll stay at my friend's apt... What I forgot is that my roommate Nacho left to go get wasted tonight for cinco de mayo... and he tends to like to bust into my room when he gets back drunk and wake me up for full conversations, should be an interesting night... – Nicolas Villanueva May 06 '11 at 04:39

I think Arturo's answer is spot on, but not complete. There is a sense in which what you did is actually true.

Take the first series:

$$\begin{align*} s_1 &= 1;\\ s_2 &= 1+2;\\ s_3 &= 1+2+4;\\ &\vdots \end{align*}$$

and the second series you obtain by multiplying the first by two

$$\begin{align*} s'_1 &= 2;\\ s'_2 &= 2+4;\\ s'_3 &= 2+4+8;\\ &\vdots \end{align*}$$

Now, if you take the following limit:

$$\lim_{n \to \infty} s'_n - s_{n+1} = -1 \; ,$$

you indeed obtain your result. However if you take the following

$$\lim_{n \to \infty} s'_n - s_n = \infty \; .$$

So, by your little manipulation, you were implicitly doing the first limit, hence your result. However, since $\infty-\infty$ is an indeterminate form, altering the limit a bit will lead to a different result.

In fact, as Qiaochu explains in his answer, there are settings in which the natural interpretation of series is precisely the one corresponding to your calculation.

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In a book called $p$-adic Numbers, by Fernando Q. Gouvea, this is problem 21 on page 20. My edition says Second Edition, Corrected 3rd printing 2003. It is in the Universitext series of the publisher Springer-Verlag, softcover. Your teachers might enjoy this book, it can be ordered from this site.

Problem 21 says: Show that, for any prime $p,$ the formula $$ 1 + p + p^2 + p^3 + p^4 + \cdots = \frac{1}{1-p} $$ is true in $\mathbb Q_p.$

What I wanted to emphasize was the partial expression $$ (1-p) \left( 1 + p + p^2 + p^3 + p^4 + \cdots + p^n \right) = 1 - p^{n+1} $$

As you know, in "ordinary" precalculus $p^{n+1}$ gets bigger and bigger as $n$ increases. But in the $p$-adic numbers $\mathbb Q_p,$ the method of measuring sizes of numbers says that the expression $p^{n+1}$ is getting smaller and smaller, and its limit is 0. The result is that, in this very special setting, it is valid to say $$ (1-p) \left( 1 + p + p^2 + p^3 + p^4 + \cdots \right) = 1 $$ Your paradox is the case $p=2.$ Note that the two occurrences of the prime $p$ must match. The limit involving the prime $p$ becomes false again in $\mathbb Q_q,$ for some prime $q \neq p.$

Will Jagy
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  • [The book in Google Books.](http://books.google.com/books?id=g1QHBOBzo9kC&pg=PA20) – J. M. ain't a mathematician May 06 '11 at 18:46
  • Thank you, J.M. GiggleBooks is a bit finicky, the first time it would not show me page 20, but after that yes. It says clearly that page 22 is not part of the preview. Let me say that I also like the short discussion of Hensel's lemma on page 211 of "Problems from the Book," by Titu Andreescu and Gabriel Dospinescu. $$ $$ http://awesomemath.org/xyz-press/ $$ $$ Andreescu is the main guy in $$ $$ http://www.artofproblemsolving.com/Forum/index.php $$ $$ MSE seems to get the occasional Olympiad type problem, and those are discussed ad nauseam at AoPS. – Will Jagy May 06 '11 at 19:10
  • Dear Will, nice to see you here! I hope you enjoy it. – Willie Wong May 06 '11 at 19:15
  • Hi, Willie. There seems to be more of a premium on matching the explanation to the OP here. Well, it took me months to get attuned to MO. – Will Jagy May 06 '11 at 19:28

Making the sum symmetric serves to simplify it so that the mistake becomes obvious, namely

$$\rm\ x\ = \sum_{k\ \in\ \mathbb Z}\ 2^{\:k}\ \ \Rightarrow\ \ 2\ x\ =\ x\ \ \not\Rightarrow\ \ x\ =\ 0$$

Bill Dubuque
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