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:
$x=-2$

$\def\bR{\mathbb{R}}\def\bQ{\mathbb{Q}}$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 $\bQ$.
