How to find closed form of a binomial series.

When working on a problem, I needed to find the closed form of the infinite sequence:

$$1 - 2x + 3x^2 - 4x^3 + \cdots$$

I struggled with this for a while and eventually found, through the Internet, that it is equal to $(1+x)^{-2}$.

How could I have approached this in such a way that I would have arrived at the answer myself?

There are many ways to get the function from this series. One is to use negative binomial coefficients: $$ \begin{align} \sum_{k=0}^\infty(-1)^k\binom{k+1}{1}x^k &=\sum_{k=0}^\infty(-1)^k\binom{k+1}{k}x^k\\ &=\sum_{k=0}^\infty\binom{-2}{k}x^k\\[4pt] &=(1+x)^{-2} \end{align} $$

Another method would be to subtract the geometric series $$ \frac1{1+x}=1-x+x^2-x^3+x^4-x^5+\dots $$ to get $$ \begin{align} f(x)-\frac1{1+x} &=-x+2x^2-3x^3+4x^4-\dots\\ &=-xf(x) \end{align} $$ Solving for $f(x)$ gives $$ f(x)=\frac1{(1+x)^2} $$

Method 1. Differentiate $$1-x+x^2-x^3+\cdots=\frac1{1+x}\ .$$

Method 2. Write the sum out this way: $$\eqalign{ 1-x+x^2-x^3+x^4-\cdots&\cr {}-x+x^2-x^3+x^4-\cdots&\cr {}+x^2-x^3+x^4-\cdots&\cr {}-x^3+x^4-\cdots&\ .\cr}$$ Adding up the first row, then the second and so on by geometric series gives $$\frac1{1+x}-\frac{x}{1+x}+\frac{x^2}{1+x}-\frac{x^3}{1+x}+\cdots$$ which is another geometric series.