Infinite sequence of $\sqrt{2^n}$ equals $i$? So I am no mathematician, in fact I consider myself not very good at math at all, however I do enjoy it. Anyways, I was messing around when I remembered a numberphile video I watched a while back about solving the infinite sequence 1 + 1/2 + 1/4 etc.
I thought, "can I apply this to the powers of two?" So, I set up an infinite sequence of the powers of two:
$$s = \{1 + 2 + 4 + 8...\}$$
When you multiply it by two...
$$2s = \{2 + 4 + 8 + 16...\}$$
Well, then you could say...
$$2s = s - 1$$
$$s = -1$$
Now I thought this was interesting, so I expanded on it:
$$\sqrt{s} = \sqrt{-1}$$
or 
$$\{\sqrt{1} + \sqrt{2} + \sqrt{4}...\} = i$$
or
$$\{\sqrt{2^0} + \sqrt{2^1} + \sqrt{2^2}...\} = i$$
This I thought was especially interesting, so I tried to convert it to an equation:
${\sum_{n=0}^{\infty}} \sqrt{2^n} = i$
(I apologize for the bad LaTeX by the way)
Anyway, all I want to know is whether I got this right or not, and if I didn't then I'd like to know why.
 A: Your first claim that $\sum_{n=0}^\infty 2^n = -1$ is incorrect, but interesting and I'll discuss it below.
In your second claim, you make the error $\sqrt{a+b} = \sqrt{a} + \sqrt{b}$. This is incorrect for the same reason $(a+b)^2 = a^2 + b^2$ is incorrect (instead $(a+b)^2 = a^2 + 2ab + b^2$).

You used:
$S = 1 + x + x^2 + \ldots$
$xS = x + x^2 + x^3 + \ldots$
$S - xS = 1$
$(1-x)S = 1$
$S = \frac{1}{1-x}$
To get
$\sum x^n = \frac{1}{1-x}$
and you applied this for $x = 2$ to get $\sum 2^n = -1$.
This sum diverges for $|x| \geq 1$, which is to say, if you add up the numbers, they don't get closer and closer to anything. For $|x| < 1$, they do get closer and closer, as you can see explicitly with $1 + 1/2 + 1/4 + \ldots$.

Plugging in $2$ for $x$ in the above formula is a type of analytic continuation. You found a formula that worked for some values of $x$, and tried it for values of $x$ that don't make sense in your original sum (because the sum diverges).
There are interesting reasons to perform analytic continuation, and a long history of it.
But the sum doesn't actually equal $-1$.
A: We should be very careful about our series convergence.
The series you have defined is $s=\sum_{n=0}^\infty 2^n=\infty$
And so, $s-1=\sum_{n=1}^\infty 2^n=\infty $. 
So the thing which u got as $s=-1$ is wrong.
Here itself answer is over. 
Extra
In the further steps u have used $\sqrt s=\sqrt{\sum 2^n}= \sqrt 1+\sqrt 2+\sqrt {2^2}+\ldots$ which is not true.
