Sum of non negative powers of 2 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?
 A: 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.
A: $$\begin{aligned}n&:=2 + 4 + 8 + 16 \cdots\\ \iff n&= 2(1 + 2 + 4 + 8 + 16\cdots) \\ \iff  n &= 2(n + 1) \\ \iff n  & = -1\end{aligned}  $$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.
A: 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: http://en.wikipedia.org/wiki/P-adic_number)
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.
