Now of course I'm not stranger to the fact that adding finite (and in many cases - infinite) amount of positive numbers always yeilds a positive number, but in many cases, often the finite limit isn't equivalent to the strange nature of infinity. It seems that mathematics tends to prefer that if $\sum_{n=0}^\infty2^n$ were to converge, it likes
$$\sum_{n=0}^\infty2^n=-1$$
This can be seen in many "proofs" like forcing the geometric series formula:
$$\sum_{n=0}^\infty r^n=\frac{1}{1-r}\implies\sum_{n=0}^\infty2^n=\frac{1}{1-2}=-1$$
Or the digit function (which returns the digit in the $b^n$ column of $x$ in base $b$):
$$D_b^n(x)=\lfloor\frac{x}{b^n}\rfloor-b\lfloor\frac{x}{b^{n+1}}\rfloor$$
(E.g; $D_{10}^{-2}(\pi)=4$)
For $-1$: $$D_2^n(-1)=1 \forall n>0$$ $$D_2^n(-1)=0 \forall n\leq0$$
Hence it could be seen that $-1_{(10)}=\bar1.0_{(2)}$, which implies $$\sum_{n=0}^\infty2^n=-1$$
Besides conventions for convergence, what exactly is stopping this sum from being equal to $-1$? Would such an equivalence lead to a contradiction?
Computationally concerned, using $-1_{(10)}=\bar1.0_{(2)}$ as a binary expansion holds all the expected arithmetic properties like $\bar1.0_{(2)}\times\bar1.0_{(2)}=\bar01.0_{(2)}$
is a true statement.
So is the statement
And so is the statement
All three statements are true because $A\implies B$ is always true if $A$ is false. The simple fact is that the expression $$\sum_{n=0}^\infty 2^n$$ denotes the sum of a divergent series, and it therefore cannot be used in an equality because it is undefined.
The expression $$\sum_{n=0}^\infty a_n$$ is simply shorthand for $$\lim_{N\to\infty} \sum_{n=0}^Na_n.$$
If you plug in $a_n=2^n$ into that expression, you see that what you are really asking is why
$$\lim_{N\to\infty} \sum_{n=0}^N 2^n=-1$$ is not true. And it is not true because the right-hand side of the equality is a real number, while the left hand side is an undefined expression.