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Possible Duplicate:
Infinity = -1 paradox

I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:

Do I have to?

Okay so, Let $x = 1+2+4+8+\dots$

$2x-x=x$

$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$

Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$.

Therefore, $1+2+4+8+\dots = -1$.

I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong?

Community
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umop aplsdn
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    The algebraic operations are not valid if the sum doesn't exist (which it doesn't). Also, some infinite series (conditionally convergent) are very sensitive to rearrangement, so the rearrangements must be valid on the level of partial sums in order to be valid for the infinite sums. However, these very same algebraic manipulations can be used to give analytic continuations which give the regularized value of $-1$ for the divergent sum. (Alternatively you can switch out the Euclidean topology for the $2$-adic topology, in which case this is completely valid.) – anon Sep 11 '12 at 01:00
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    If your friend had told you that this sum was $-\frac{1}{12}$.. now that would've been much more interesting! – student Sep 11 '12 at 01:04
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    (Yes, there is another divergent series $1+2+3+\cdots$ whose zeta-regularized value is $-1/12$.) – anon Sep 11 '12 at 01:05
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    Quite relevant: Divergent series and $p$-adics – MJD Sep 11 '12 at 01:48
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    +1 MJD. This series converges (to -1) in the 2-adic metric. – GEdgar Sep 11 '12 at 01:51
  • Its interesting to note that if you subtract 1 from each side you will get 2+4+8+... = -2 And repeating in this process you get n+ (n+1) +... = -n and at infinity, Infinity = -Infinity. Its really interesting to note that Ramanujan had an entire section in one of his notebook on what such summations "converge" to. I think he called it "theory of divergent series" or something along those lines... interesting games! – Squirtle Sep 11 '12 at 04:26
  • Wikipedia has an article about this series: http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_... – Michael Hardy Sep 19 '12 at 15:43

5 Answers5

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The first mistake is right at the beginning, in writing "Let $x=1+2+4+\cdots$." This builds in the assumption that there is such an object as $1+2+4+\cdots$. The second mistake lies in treating this supposed object as if it were a finite but maybe very long sum, to which the sensible rules for manipulating finite sums apply.

Remark: Think about the "sum" $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$. If we call this $x$, and use a manipulation analogous to the one that you made, we end up with the conclusion that the "sum" is $2$. When infinite series are formally defined, it turns out that the answer is indeed $2$. So some "natural" manipulations yield nonsense, and some yield correct results. Of course that is not a tolerable state of affairs: we cannot use manipulational techniques that sometimes yield a correct result, and sometimes don't. This sort of issue, at a more sophisticated level, led mathematicians in the second half of the $19$th century to look for very careful definitions of the fundamental objects of mathematics, and rigorous proofs of their basic properties.

André Nicolas
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  • And if the sum is finite, say length $n$, the proof is flawed anyway, as the multiplication of the LHS sum by two yields a term of the sum that doesn't exist in the RHS, namely $2^n$ - so the two sums do not fully cancel out and the net result is $2^n - 1$ which is the expected result. – Thomas Sep 11 '12 at 02:41
  • While true, I consider this answer incomplete, since there are generalized summations that allow these kinds of manipulations even for divergent series. And even when those do not work, one has further methods like zeta-regularization and p-adic interpretation. Or in other words, there are many definitions of summation (and arguably much more useful than the standard one in many situations). – Marek Sep 11 '12 at 09:51
  • @Marek: Yes, it is incomplete. My aim was to present a short fairly vigorously expressed answer that would be locally useful. – André Nicolas Sep 11 '12 at 14:39
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The infinite series $\displaystyle 1 + 2 + 4 + 8 + \ldots$ diverges. However, the sum $f(z) = 1 + z + z^2 + z^3 + \ldots$, which converges to $1/(1-z)$ for $|z| < 1$, has an analytic continuation to the complex plane with the point $1$ removed, and indeed $f(2) = -1$. So in that sense you could regard $-1$ as the value of the divergent series.

For more on this, see http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7 and references there.

Robert Israel
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Look at Calculus textbook, undergraduate level. When we treat infinite sum, we cannot change the order to compute.

Example. $1-1+1-\cdots$

$$(1-1)+(1-1)+\cdots=0+0+\cdots=0$$

$$1+(-1+1)+(-1+1)+\cdots=1.$$

Michael Hardy
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Tom
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  • That is conditional convergence, not relevant in this case, for example $1-\frac{1}{2^2}+\frac{1}{3^2}-\frac{1}{4^2}+..$ you can change the order. – jimjim Sep 11 '12 at 05:47
  • @Arjang: True, but the naive mistake is the same - that you can do whatever to an infinite sum and everything is fine. – user1729 Sep 11 '12 at 09:15
  • @user1729 : everything is fine, what is not fine is the naive use of convergence where the more complete forms of convergence need to be considered, cesaro summibility etc.. And this example converges using extended version of convergence. – jimjim Sep 11 '12 at 11:24
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    @Arjang: But surely that is the point - that naive convergence doesn't work. – user1729 Sep 11 '12 at 12:36
  • What I wanted to say is: in the post, between "Therefore" and "Now", he changed the order of sum, which is the point of mistake. – Tom Sep 19 '12 at 13:30
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It is similar to "proving" that 1 = 2 by saying that 1+infinity = 2+infinity.

vsz
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What is wrong is that ``$1+2+4+8+\cdots$'' is not a number, and so you cannot treat it like one.

Martin Argerami
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