Math riddle: playing with infinite series

Since I haven't had that much time to work on new posts (there's also a follow up on the previous post already that I hope I can publish soon), I thought I show you a math riddle that I find quite amusing and interesting because the calculations beeing made here are completely valid (no hidden 0-divisions or tricks of that sort). I've shown this to a couple of mathematicians and they have also been surprised.

So, without further ado, here it is:


As we may intuitively think, the statement about this series seems accurate:
$$1+2+4+8+16+...=\sum _{i=0}^{\mathrm{\infty }}{2}^i=\mathrm{\infty }$$
...but is it really $\mathrm{\infty }$? Let's break it down: we can multiply the same sum by 1 without altering the result, so
$$1+2+4+8+16+...=1\times (1+2+4+8+16+...)$$
Since $1=2-1$, we can replace it in the equation above like so:
$$1\times (1+2+4+8+16+...)=(2-1)\times (1+2+4+8+16+...)$$
Pretty basic right? Let us now apply the distributive property to get this:
$$(2-1)\times (1+2+4+8+16+...)=2\times (1+2+4+8+16+...)-1\times (1+2+4+8+16+...)$$
...and if we get rid of the parenthesis we'll get something like this:
$$=2+4+8+16+32+...-1-2-4-8-16-32-...$$
You may have noticed already that all the numbers in the left, cancel out with all numbers in the right, except for a remaining -1:
$$\text{⧸}2+4+8+16+32+...-1-\text{⧸}2-4-8-16-32-...$$
$$\text{⧸}2+\text{⧸}4+8+16+32+...-1-\text{⧸}2-\text{⧸}4-8-16-32-...$$
$$\text{⧸}2+\text{⧸}4+\text{⧸}8+16+32+...-1-\text{⧸}2-\text{⧸}4-\text{⧸}8-16-32-...=-1$$
Since we come from a series of equalities, we can safely conclude that:
$$\sum _{i=0}^{\mathrm{\infty }}{2}^i=-1$$
Fun, isn't it?

If the intuitive notation for the series $(1+2+4+8+...)$ bothers you, you can replace it by any name you want, lets say $S$:
$$1+2+4+8+...=\sum _{i=0}^{\mathrm{\infty }}{2}^i=S$$

An alternative way to get the same result is to take the common factor 2 out of the series (starting from the second term of course) like so:
$$S=1+2+4+8+16+...=1+2(1+2+4+8+...)=1+2S$$
and from here is just a matter of letting  the $S$ on both sides cancel out and you'll get a nice contradiction:
$$S=1+2S$$
$$2S-S=-1$$
$$S=-1$$
...and that's how it's done.