What if $x$ is, say, the number $2$?

Then the geometric series formula says that $1+2+4+8+16+...$ equals $\dfrac {1} {1-2}$, which is $-1$. That’s just absurd!

On the other hand, if $x = 0.1$, then the geometric series formula says that

$1+\left(0.1\right)+\left(0.1\right){2}+\left(0.1\right){3}+... = 1+0.1+0.01+0.001+...=1.111...$

equals

$\dfrac {1} {1-0.1}=\dfrac {1} {0.9}=\dfrac {10} {9}$.

This is one and one-ninth. From primary school we learn that $\dfrac {1} {9}$ as a decimal is $0.111...$. (Also see the Decimalia Island!)
So one-and-one-ninth equals $1.111...$. The geometric series formula is correct in this case!

$1+\left(0.1\right)+\left(0.1\right){2}+\left(0.1\right){3}+... = \dfrac {1} {1-0.1}$

What is going on? When can we believe the formula and when can we not?

ALGEBRA VERSUS ARITHMETIC

The $1\leftarrow x$ machine is a machine that displays mechanical algebra. Everything it does and all we conclude from it is true in terms of mechanical algebra.

For example, the geometric series formula says that $1+2+4+8+...$ equals $\dfrac{1}{1-2}$, which is absurd as a statement of arithmetic. However, in a purely mechanical sense, without regard to arithmetic, there is some version of truth to this claim. Here’s how we can see some truth.

If $1+2+4+8+...$ equals $\dfrac{1}{1-2}$, then multiplying $1+2+4+8+...$ by $1-2$ should give $1$. Does it?
Yes it does!

$\begin{aligned}$
\left(1-2\right) \times \left(1+2+4+8+...\right) &= \left(1-2\right) + \left(1-2\right) \times 2 + \left(1-2\right) \times 4 + \left(1-2\right) \times 8 + ...\
&= 1-2+2-4+4-8+8-16+...\
&= 1+0+0+0+...\
&= 1
\end{aligned}

So $1+2+4+8+...$ really does behave like $\dfrac{1}{1-2}$.

But still, this is not very satisfying. We want to know when $1+x+x{2}+x{3}+... = \dfrac {1} {1-x}$ is actually true as a statement of arithmetic.

THE CALCULUS ANSWER (FOR THE BOLD)

One studies infinite sums in a calculus class. You learn there, for example, that $1+x+x{2}+x{3}+... = \dfrac {1} {1-x}$ is true as a statement of arithmetic for small values of $x$, (specifically, for all values of $x$ strictly between $-1$ and $1$.) The formula is valid for $x=0.1$, as we saw, and not for $x=2$.

For those who are game, here is how the calculus argument goes.

Regular polynomial division shows that $\dfrac {1-x^2} {1-x}=1+x$ and $\dfrac {1-x^3} {1-x} = 1+x+x^{2}$ and $\dfrac {1-x^4} {1-x} = 1+x+x{2}+x{3}$, and so on. (Try these!)
In general, we see that
$1+x+x{2}+...+x{n-1}=\dfrac {1-x^n} {1-x}$.

Now as we let $n$ get bigger and bigger it looks like we’re getting the infinite geometric sum.
$\begin{aligned}$
1+x &= \dfrac{1-x^2}{1-x}\
1+x+x^2 &= \dfrac{1-x^3}{1-x}\
1+x+x^2+x3 &= \dfrac{1-x^4}{1-x}\
\vdots\
1+x+x^2+x3+... &= ?
\end{aligned}

So the question is: What does $\dfrac {1-x^n} {1-x}$ go to as $n$ gets bigger and bigger? If there is an answer to this question, then the answer will be the value of $1+x+x{2}+x{3}+...$.

So does $\dfrac {1-x^n} {1-x}$ have a limit value? Well, this depends on whether or not $x^{n}$ has a limit value as $n$ grows. So for which values of $x$ do the powers of it approach a value?

We know the that powers of $0.1$, for example, and of $0.83$ and of $- \dfrac {1}{2}$, all approach zero for bigger and bigger powers. In fact, $x^{n}$ gets closer and closer to zero as $n$ grows for any value between $-1$ and $1$.

So for $-1<x<1$, we have
$\begin{aligned}$
1+x &= \dfrac{1-x^2}{1-x}\
1+x+x^2 &= \dfrac{1-x^3}{1-x}\
1+x+x^2+x3 &= \dfrac{1-x^4}{1-x}\
\vdots\
1+x+x^2+x3+... &= \dfrac{1-0}{1-x} = \dfrac{1}{1-x}
\end{aligned}

The geometric series formula can be believed, as a statement of arithmetic, for $-1<x<1$, at least.

THE HONEST APPROACH

Another approach to examining the geometric series formula is to make an explicit leap of faith. Let’s assume that the infinite sum $1+x+x{2}+x{3}+...$ is meaningful and has an answer. Call that answer $S$.

Then

$\begin{aligned}$
S &= 1+x+x^{2}+x{3}+...\

&= 1+x \left(1+x+x^{2}+...\right) \
&= 1+xS
\end{aligned}
from which we get $S= \dfrac{1}{1-x}$.

This is an honest approach. It proves:
IF the infinite sum $1+x+x^{2}+...$ has an answer, then that answer must be $\dfrac{1}{1-x}$. It makes no assertion as to whether or not the infinite sum is meaningful in the first place.

The same is true for the dots-and-boxes approach as it too proves: IF $1+x+x{2}+x{3}+...$ is meaningful to you, then it has the answer $\dfrac{1}{1-x}$. It is up to you to decide if this infinite sum is meaningful.
(Calculus likes to say it is if $-1<x<1$.)

OTHER SYSTEMS OF ARITHMETIC OFFER OTHER MEANINGS

The statement $1+2+4+8+...=-1$ is meaningless in ordinary arithmetic. But who says we have to stay with ordinary arithmetic? Is there an extraordinary way to view matters?

We tend to view numbers as spaced apart on the number line additively. Walk one step to the right of $0$ and we end up at position $1$. Now add to that two steps and we end up position $3$. Now add four steps, position $7$. And so on. The sum $1+2+4+8+...$, in this viewpoint, takes infinitely far to the right of $0$ on the number line.

$1+2+4+8+...=\infty$ in ordinary arithmetic. It does not equal $-1$.

But let’s think of numbers multiplicatively. In particular, since we are focusing on the sum $1+2+4+8+...$, let’s think of factors and multiples of powers of two.

Now $0$ is a highly divisible number. It is the most divisible number of all. With regard to just two-ness it can be divided by $2$ once, in fact twice, in fact thrice. In fact, you can divide $0$ by two as many times as you like—and still keep dividing.

With regard to two-ness, the number $8$ is somewhat zero-like: you can divide it by two three times. But $32$ is even more zero-like: you can divide it by two five times. And $2^{100}$ is even more zero-like still.

So in this sense, $2^{100}$ is a number very close to $0$. The number $32$ is somewhat close to $0$. The number $8$ is less close. The number $1$ is not very close to zero at all: it cannot be evenly divided by two even once.

So, in this context, could $1+2+4+8+...$ possibly be $-1$?

Well

$\begin{aligned}$
1+2 = 3 &= 4-1\
1+2+4 = 7 &= 8-1\
1+2+4+8 = 15 &= 16-1\
\vdots &= \vdots \
1+2+\ldots + 2^{99} &= 2^{100}-1
\end{aligned}

These finite sums grow to become “a number very close to zero, minus one.” In the limit, the infinite sum thus has value $0-1 = -1$, just as our formal arguments said it would be.

So in this multiplicative view of arithmetic, $1+2+4+8+...$ is a meaningful quantity, and it does indeed have value $-1$. The geometric series formula is meaningful and correct for $x = 2$ in this context.