Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$ See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$
$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
Then:
$x=1$
Now I use the same argument to prove $2+4+8+...=-2$
$x=2+4+8+...$
$2x=4+8+16+...$
Then:
$x=-2$
 A: $\def\bR{\mathbb{R}}\def\bQ{\mathbb{Q}}$Well, $\sum_{k\geq1} 2^k$ converges to $-2$, but not in $\bR$, it does so in $\bQ_2$, the $2$-adic completion of $\bQ$. (In $\bR$, it obviously diverges.)
Let me prove that $\sum_{k\geq0} 2^k\to-1$, which is the same (just multiply by $2$). We have, for the partial sums, $A_n:=\sum_{k=0}^{n-1} 2^k= 2^n-1$ and then $A_n-(-1)=2^n$ and $2^n\to0$ in the $2$-adic sense.
So what you discovered is that series that are divergent in $\bR$ need not be divergent in other completions of $\bQ$.
A: Hint the series converges only if $|r|<1$ so your second proof is wrong as $|r|=2$
A: The proof does work if you interpret it correctly.
In the proof for $\frac12 + \frac14 + \cdots = 1$, you are using the fact in
$\mathbb{R}$, the successive term $\frac{1}{2^n}$ is getting smaller and smaller as $n$ getting bigger and bigger.
The partial sums
$$\sum_{n=1}^{N} \frac{1}{2^n} = \frac{\frac12 - \frac{1}{2^{N+1}}}{1 - \frac12} = 1 - \frac{1}{2^N}$$
getting closer and closer to $1$ as $N$ increases. This means in $\mathbb{R}$, we have
$$\sum_{n=1}^\infty \frac{1}{2^n} = \lim_{N\to\infty} \sum_{n=1}^N \frac{1}{2^n}
\text{ exists and equal to } 1$$
If you look at the other sum $2 + 4 + \cdots$, the successive term $2^n$ doesn't getting smaller and smaller as $n$ increases. The corresponding partial sums doesn't converge. In order for your argument to work, a prerequisite is the partial sum converges. When the partial sums fail to converge, the difference of two indeterminates is an indeterminate. Your argument will fail
and you cannot conclude $2 + 4 + \cdots = -2$ when working within $\mathbb{R}$.
However, there is more than one way to extend $\mathbb{Q}$. For each prime number $p$, there is a beast $\mathbb{Q}_p$ called $p$-adic numbers. In particular, for
$p = 2$, the successive terms $2^n$ does getting smaller and smaller as $n$ increases. If you work within the $2$-adic numbers $\mathbb{Q}_2$, you argument will work and the series does converge to $-2$ there.
A: You can't subtract divergent series.
A: In the proof, you have begun by implicitly saying "Assume that the sequence converges, and call the limit $x$". For $1/2 + 1/4 + 1/8 + \cdots$, you obtain $x = 1$. For $2 + 4 + 8 + \cdots$, you obtain $x = -2$.
But that means you have only proved "If $2 + 4 + 8 + \cdots$ converges, then it converges to $-2$". This does not show that the series $2 + 4 + 8 + \cdots$ actually converges. That has to be established separately. 
In the case of $1/2 + 1/4 + 1/8 + \cdots$, you can prove by induction that the partial sums are bounded by $1$, and then use the monotone convergence theorem to prove that the series converges. Until you prove the series converges, you don't actually know from your previous calculation that the limit of $1/2 + 1/4 + 1/8 + \cdots$ is $1$. You only know that if $1/2 + 1/4 + 1/8 + \cdots$ converges, then the limit is $1$.  
Now, you can't prove the series $2 + 4 + 8 + \cdots$ converges, because it doesn't. But you can prove that if it did converge the limit would be $-2$ - which is still consistent with the series not converging.  This is all that your calculation shows.
So the method you are using is fine, but you have to remember that it is relatively meaningless to try to compute the value of a series if the series doesn't converge. When you use the method in the question to compute the value of a series, you have to prove separately that the series converges. 
