I solved this equation something like this, as shown in the photo:
Is it correct?
If I put $x=2$ I get weird results!
Answer
As a lot of the comments and the other answers point out, this only works for $-1
$$
-1 - \frac 1x - \frac{1}{x^2} - \frac{1}{x^3} - \cdots
$$
which is a sequence that converges as long as $|x|> 1$, and it gives $\frac{x}{1-x}$ (or really, $\frac{1}{1/x-1}$, which ammounts to the same thing) if you do the same trick as you've done in your question. This is what is called "the series expansion of $\frac x{1-x}$ around $\infty$" (since the series converges as long as $x$ is large enough), while $x + x^2 + x^3 + \cdots$ is the series expansion around $0$ (since it converges as long as $x$ is close enough to $0$ [there is a bit more to it than that, but that's details]). Together, they give you geometric series that evaluate to $\frac{x}{x-1}$ on the whole number line except at $-1$ and $1$.
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