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I found this very interesting. I looked at the part where it mentioned about the uppasala lectures on calculus. It mentions that in the first lecture it was proved that the sum of all the non negative powers of two equals $-1$ and even this equality was demonstrated in a practical way.

$2+4+8+16+32\cdots$ in no way seems to be summing up to $-1$. All are positive numbers, so the sum must be positive. Can I get an explanation for this?

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It must be something like this – Guest 86 Jan 22 '13 at 15:37
Can you be more specific on the link? Your link has a lot of stuff. – Patrick Li Jan 22 '13 at 15:39
Since the partial sums do not tend to 0, the sum $\sum_{i=1}^{\infty} 2^i$ does not converge.... until one changes the underlying notion of distance. In the 2-adic topology, in which numbers are "close" when they differ by a large power of 2, the sum makes perfect sense, and converges to $\cdots 1111110$, which is 2-adic equivalent to $-2$ (you have to include $2^0$ in the sum to get $-1$). – Shaun Ault Jan 22 '13 at 15:45

The standard 'proof' looks like this. Let $$S = 1 + 2 + 4 + 8 + 16 + \cdots = \sum_{n=0}^{\infty} 2^n$$ be the sum of the positive powers of $2$. Then $$2S = 2 + 4 + 8 + 16 + 32 + \cdots$$ so we have

$$S = (2S-S) = (2+4+8+\cdots) - (1+2+4+8+\cdots) = -1$$

However this is invalid. The equation $\sum a_n \pm \sum b_n = \sum (a_n \pm b_n)$ only holds when both $\sum a_n$ and $\sum b_n$ converge. If they don't converge then it breaks down because it amounts to trying to give $\infty-\infty$ a well-defined value.

That said, there are times when it might be useful to consider the sum to be equal to $-1$. For more on this, see the Wikipedia article $p$-adic numbers (specifically, $2$-adic numbers). Also this article.

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$$\begin{aligned}n&:=2 + 4 + 8 + 16 \cdots\\ \iff n&= 2(1 + 2 + 4 + 8 + 16\cdots) \\ \iff n &= 2(n + 1) \\ \iff n & = -1\end{aligned} $$The above is your "proof". Now, the problem is how the expressions diverge. We can not simply add or subtract two expressions where one is convergent and the other is divergent; both, in that case, will be divergent and you'll end up generating another fake proof.

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One serious and one less serious solution

(1) If one says that a series is divergent then the one has to define in what sense it diverges. The norm referred to by the answerers above was the absolute value of the integers in question. There are other norms over the field of rationals. We are going to examine the convergence problem at stake assuming the so called $2$-adic norm. (see:

Every rational $q$ can be uniquely written as $$q=2^{\ r}\frac{m}{n},$$

where $m,n$ are relative primes and $2$ does not divide either of them. The so called $2$-adic norm of $q$ is then, by definition, $$|q|_2=2^{-r}.$$ (Show that this is a norm...) Now $$\lim_{n\rightarrow \infty}|\sum_{i=0}^n2^{\ i }-(-1)|_2=lim_{n\rightarrow \infty}\ 2^{-(n+1)}=0.$$ That is, we have a convergent series.

(2) In two's complement notation $1111...111$ means $-1$. One meaning of this sequence is the series above and the other meaning is $-1$ and this is independent from the number of digits we use.

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Correction, when one works with a different topology (or a norm inducing one, what have you) rather than the standard topology, then one has to state that clearly. When one defines a different sense for convergence and divergence than the usual one, then one has to state that explicitly. When one states nothing, it is seemingly assumed that they are using the standard topology and norm and definition of convergence. – Asaf Karagila Feb 11 '15 at 13:06
Our statement does not make any sense in one topology and it does in another topology. Considering solution (2) above, the intuitive content of the statement is independent of the topology chosen. That is, we are not talking about two different worlds. We rather don't know exactly what we are talking about. Why do we have to stick to foundations? I deal with this problem only because of philosophical considerations. – zoli Feb 11 '15 at 13:22

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