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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 $x = 1+2+4+8+\dots$


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

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?

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marked as duplicate by Argon, Austin Mohr, William, Jyrki Lahtonen, m_t_ Sep 11 '12 at 10:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 1:00
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 1:04
(Yes, there is another divergent series $1+2+3+\cdots$ whose zeta-regularized value is $-1/12$.) – anon Sep 11 '12 at 1:05
Quite relevant: Divergent series and $p$-adics – MJD Sep 11 '12 at 1:48
+1 MJD. This series converges (to -1) in the 2-adic metric. – GEdgar Sep 11 '12 at 1:51
up vote 19 down vote accepted

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.

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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 2:41
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 9:51
@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

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 and references there.

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

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



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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. – Arjang Sep 11 '12 at 5:47
@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 9:15
@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. – Arjang Sep 11 '12 at 11:24
@Arjang: But surely that is the point - that naive convergence doesn't work. – user1729 Sep 11 '12 at 12:36
Thanks user1729. – Tom Sep 19 '12 at 13:28

It is similar to "proving" that 1 = 2 by saying that 1+infinity = 2+infinity.

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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.

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@Arjang: I think Euler is dead for quite some time now. – Asaf Karagila Sep 11 '12 at 9:20
@AsafKaragila : His works make him more alive than many alive people. – Arjang Sep 11 '12 at 11:26
@Arjang: I think you and modern science have very different definition of "alive". – Asaf Karagila Sep 11 '12 at 12:04
@Arjang: the fact that he was genius doesn't preclude that he did make mistakes. – Martin Argerami Sep 11 '12 at 20:24
@MartinArgerami But not there. – Did Oct 30 '13 at 12:40

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