Is it true that ${1\over x}=1-x+x^2-x^3+\dots$? Is it true that ${1\over x}=1-x+x^2-x^3+\dots$?
It was used during a formal solution regarding big $O$ notation as if it were a fairly familiar identity, but there was nothing mentioned about area of definition and so. I myself have progressed relatively enough to have encountered that expansion at some point, I believe. Something in that equality seems somewhat confusing, and I can't seem to arrive at it using Taylor Expansions. It takes some ugly work to find a pattern of an expansion around a nonzero number.
Could you should shed some light on this please? 
(I am sorry for all the tags: if I knew what area gives a simplified background to that identity, if true, I would tag it respectively.)
 A: HINT : Take $x=0$. Is the expression equal to $\dfrac {1}{0}$ ???

 It is essentially a geometric series with common ratio $r=-x$. For $|r| > 1$, the series diverges and thus the sum is infinity $\Bigg($ and NOT $\dfrac{1}{x}$ $\Bigg)$. The formula for sum of an infinite geometric series with $|r|<1 $ is $\dfrac{a}{1-r},$ where $a$ is the first term of the series. In this case, $a=1$, and $r=-x$. Therefore the infinite sum is $\dfrac{1}{1-(-x)} = \dfrac{1}{1+x}$. Therefore, concluding the answer, $$\\\boxed {\sum \limits_{k=0}^{\infty}(-1)^kx^k = \dfrac{1}{1+x} \,\,\,\,\,\, for \,\,\,\,\,|-x|<1.}$$

A: Take $x=-0.1$.
Then $$\frac1x=-10,$$
and
$$1-x+x^2-x^3+x^4-x^5+\cdots=1.11111\cdots=\frac{10}9.$$
A: No, as others have pointed out, the real identity is
$$\dfrac 1{x+1}=1-x+x^2-x^3+x^4-\&\text c$$
You can prove this by synthetically dividing $1/(x+1)$ like below:
$$\begin{array}{c |c c} & 1 & & 0 & & 0 & & 0 & & 0 & & 0 & & 0 & \cdots\\ -1 & & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & \cdots\\\hline & 1 & & -1 & & 1 & & -1 & & 1 & & -1 & & 1 & \cdots\end{array}\tag{1}$$
Where the top row, the $1,0,0,0,0\cdots$ is the numerator in terms of $x$, and the $-1$ is the negated denominator with the $x$ value omitted.
Therefore, it follows that$$\dfrac 1{x+1}=1-x+x^2-x^3+x^4-x^5+\&\text{c}=\sum\limits_{r=0}^{\infty}(-1)^rx^r\tag{2}$$
However, note that this only converges if $|x|<1$.
A: It's  $\dfrac1{1+x}$, not $\dfrac1x$, and it's valid for $\lvert x\rvert<1$. This results from the following high-school identity
$$1-x^n=(1-x)(1+x+x^2+\dots x^{n-1}),$$
which can be re-written as $$\frac1{1-x}=1+x+x^2+\dots x^{n-1}+\frac{x^n}{1-x}.$$
If $\lvert x\rvert<1$, $x^n\to 0$ as $n\to\infty$, which proves 
$$\frac1{1-x}=1+x+x^2+x^3+\dotsm$$
Setting $x\leftarrow -x$, you obtain
$$\frac1{1+x}=1-x+x^2-x^3+\dotsm$$
