Series expansion for $x$, when $x$ is small Suppose that we are given the series expansion of $y$ in terms of $x$, where $|x|\ll 1$. For example, consider $$y=x+x^2+x^3+\cdots\qquad\qquad\qquad (1).$$ From this I would like to derive the series expansion of $x$ in terms $y$. Given that $|x|\ll 1$ we can take $y=x+O(x^2)$ such that $x\sim y$. 
It then follows from $(1)$ that $$x=y-(x^{2}+x^{3}+\cdots)\qquad\qquad\qquad (2).$$ Given that $x\sim y$, $(2)$ becomes $$x=y-y^2-y^3+\cdots\qquad\qquad\qquad (3).$$
While the first two terms of $(3)$ are correct I suspect that the remaining terms are incorrect. 
For instance given that $y=x+O(x^2)$ implies that $y^2=x^2+2xO(x^{2})+(O(x^{2}))^{2}$. Therefore, $2xO(x^{2})$ will yield a $y^{3}$ term in $(3)$ when computing $x^{2}$ in $(2)$. 
How do I find the correct term containing $y^{3}$ in $(3)$, and so on.
 A: you have a geometric series. this has a simple recipe,
$$ y = \frac{x}{1-x} $$
for $|x| < 1,$ that is $-1 < x < 1.$
As it is a Mobius transformation, we can easily invert it,
$$ x = \frac{y}{1 + y} $$
A: You have 
$$y=\sum_{n\ge 1}x^n=\frac{x}{1-x}\;,$$
so $(1-x)y=x$, and hence
$$x=\frac{y}{1+y}=y\sum_{n\ge 0}(-1)^ny^n\;.$$
A: This is not an answer but it is too long for a comment.
Just as Will Jagy commented, you can inverse the series building one coefficient at the time. This is a quite tedious task but it is doable.
Suppose that you have 
$$y=a_0+a_1+a_2x^2+a_3x^3+O\left(x^4\right)$$, replacing $x$ by $b_0+b_1+b_2y^2+b_3y^3$, replacing, expanding and identifying one coefficient at the time,  you will then arrive to $$x=\frac{1}{a_1}(y-a_0)-\frac{a_2 }{a_1^3}(y-a_0)^2+\frac{2 a_2^2-a_1
   a_3 }{a_1^5}(y-a_0)^3+O\left((y-a_0)^4\right)$$
If we set all the $a_i$'s equal to $1$ as in the example in the post, you will then have $$x=y-y^2+y^3+O\left(y^4\right)$$ If we replace $x$ by $y-y^2+y^3$ in $y=x+x^2+x^3$, we should get $$y=y+4y^5-6y^6+\cdots$$ which shows the error.
A: This topic is known as
"inversion of power series".
Do a search -
you will get many interesting hits
(it is not quite as bad as tvtropes).
A very good reference
is "generatingfunctionology"
which can be downloaded for free
here:
https://www.math.upenn.edu/~wilf/DownldGF.html
A: If we assume that the series continues as $y=\sum\limits_{k=1}^\infty x^k$, we have
$$
y=\frac{x}{1-x}\tag{1}
$$
then we can compute
$$
x=\frac{y}{1+y}=\sum_{k=1}^\infty(-1)^{k-1}y^k\tag{2}
$$
However, if we don't make this assumption, we can invert the series using Lagrange Inversion:
$$
\begin{align}
\left[\,y^k\,\right]x
&=\frac1k\left[\,x^{-1}\,\right]y^{-k}\\
&=\frac1k\left[\,x^{-1}\,\right]\left(x+x^2+x^3+\dots\right)^{-k}\tag{3}
\end{align}
$$
where $\left[x^k\right]f$ is the coefficient of $x^k$ in the power series expansion for $f$.
Since
$$
\begin{align}
\left(x+x^2+x^3+\dots\right)^{-1}&=\frac{\color{#C00000}{1}}x-1+0x+\dots\\
\left(x+x^2+x^3+\dots\right)^{-2}&=\frac1{x^2}\color{#C00000}{-}\frac{\color{#C00000}{2}}x+1+\dots\\
\left(x+x^2+x^3+\dots\right)^{-3}&=\frac1{x^3}-\frac3{x^2}\color{#C00000}{+}\frac{\color{#C00000}{3}}x+\dots
\end{align}\tag{4}
$$
$(3)$ gives
$$
x=y-y^2+y^3+\dots\tag{5}
$$
