Let $x = 1+1+1+1+1+1 ...$
Let $y=1-1+1-1+1-1 . . .$
First, let's find the value of $y$.
The partial sums of $y$ are $s_n=(1,0,1,0,1,0,...)$
If you take the means of the partial sums, you will get the sequence $(\frac11,\frac12,\frac23,\frac24,\frac35,...)$ which obviously converges to $\frac12$.
Another way (This time algebraically):
$$y=1-1+1-1+1-1 . . .$$
$$1-y=1-(1-1+1-1+1-1+1-1 ...)$$
$$1-y=y$$
$$2y=1$$
$$y=\frac12$$
The value of $y$ is therefore $\frac12$
$$x-y= (1+1+1+1+1+1+1 . . .)-(1-1+1-1+1-1+1-1 . . .) $$ $$x-\frac12 = (1+1+1+1+1+1+1 . . .)-(1-1+1-1+1-1+1-1 . . .) $$
$$x=(1+1+1+1+1+1+1 . . .)$$
$$-y=-(1-1+1-1+1-1+1 . . .)$$
So $x-y=$
$$(1+1+1+1+1+1+1 . . .)$$
$$(-1+1-1+1-1+1-1 . . .)$$
If you add those together, you'll see that the $-1$'s cancel out very other $1$, and that the $+1$'s add together, which equals:
$$0+2+0+2+0+2+0+2 ...$$ Or,
$$2+2+2+2+2+2+2 ...=2(1+1+1+1+1+1+1)=2x$$ Therefore, $x-y=2x$
$y=\frac12$
$x-\frac12=2x$
Subtract $x$ from both sides, and you get,
$$-\frac12=x$$
Is this a correct proof that $1+1+1+1+1+1 ...=-\frac12$?
$$1+1+1+\cdots$$ will always diverge to infinity. You cannot do simple series manipulations, such as reordering or adding 0, with divergent series if you are looking for an analytical value. The proper way to do this is with Zeta regularization.