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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.)

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    No. It's $\dfrac1{1+x}$, and it's valid for $\lvert x\rvert <1$. It results from a well-known high-school identity, the factorisation of $1-x^n=(1-x)(1+x+x^2+\dots+x^{n-1})$. – 2017-02-17
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    Try $\frac 1{1+x}$ - if you multiply the right-hand side of your expression by $1+x$ you will see how things cancel. It may be valid in various contexts (e.g. formal power series). – 2017-02-17
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    I thought I was wrong assuming that. Spent way too much time on that. Thank you both – 2017-02-17
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    To expand on what Bernard said, the equality would hold if you replace all of the $x$'s on the right hand side with $(x-1)$. – 2017-02-17
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    Yes this is that exact substitution I have just made. – 2017-02-17
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    @Bernard Post your comment noting the essential typo as the answer. The other posted answers address the literal question asked but not the one the OP needs an answer to. – 2017-02-17
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    Plug in 1 and see – 2018-08-27

4 Answers 4

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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 $$\frac1{1+x}=1-x+x^2-x^3+\dotsm$$

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    I like your $x \leftarrow -x$ approach ... simple and easy . :-) – 2017-02-17
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    Shouldn't it be $|-x| < 1$, as the common ratio is $-x$ ? – 2017-02-18
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    @Nirbhay: $|x| = |-x|$, therefore $|x|<1\iff|-x|<1$ – 2017-02-18
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    @celtschk Thats obviously ture .... thanks for telling ...... however the best way to write to avoid confusion is to write $|-x|$...isn't it? – 2017-02-18
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    @Nirbhay: No. I don't see what confusion would be avoided by writing $|-x|$ instead of $|x|$. – 2017-02-20
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    @celtschk Confusion = Confusion for beginners that common ration $r=|-x|$.....it is worthless to discuss over this topic....so I think we should end this discussion here only. :-) – 2017-02-21
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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, $$\\\boxed {\sum \limits_{k=0}^{\infty}(-1)^kx^k = \dfrac{1}{1+x} \,\,\,\,\,\, for \,\,\,\,\,|-x|<1.}$$

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    Note that the capitalized word "not" has been used to emphasize and not as sth offensive. ;-) – 2017-02-21
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Take $x=-0.1$.

Then $$\frac1x=-10,$$

and

$$1-x+x^2-x^3+x^4-x^5+\cdots=1.11111\cdots=\frac{10}9.$$

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    Downvote was not from me .... I appreciate your answer but I can not upvote as I have reached the daily vote limit. – 2017-02-17
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    @Nirbhay: your answer is much better ! – 2017-02-17
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    Thanks a lot for the compliment. :-) I am just a newbie at answering.. – 2017-02-17
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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 $|x|<1$.