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.
Let's kick it old school, like Euler - by which I mean blindly manipulate infinite series to reach correct solutions.
Okay, we have the series: $$A=1-2x+3x^2-4x^3+\ldots.$$ then we think, hey, all those increasing terms are annoying. Notice that $Ax$ is basically shifts the coefficients of $A$, and, since the signs alternate, adding each coefficient to the one after it is going eliminate that annoying increasing-nature of that sum and leave just some residue about the common differences.
So, let's set: $$B=A+Ax=1-x+x^2-x^3+x^4+\ldots.$$ We see that we have an alternating pattern of coefficients of $1$ and $-1$ - and hey, doing the same thing as before, we can get rid of all the coefficients by adding each coefficient to the next! So, let's do the same thing: $$B+Bx = 1-0x+0x^2+\ldots = 1.$$ That's a constant! We can work with a constant! So, how did we get there? Well, $$B=(1+x)A$$ $$1=(1+x)B$$ so $$1=(1+x)^2 A$$ implying$$\frac{1}{(1+x)^2}=A.$$
What's the takeaway? Well, if you have some funky series, you can try to reduce it by looking at patterns and trying to simplify them (esp. by taking differences between terms - we can use similar methods to relate this to more general generating functions). And if you have an identity you want to prove, think about what defines that number - $\frac{1}{(1+x)^2}$ is the number which, when multiplied by $(1+x)^2$, yields one.
(Okay, all these manipulations are pretty easy to justify as we're working with absolutely convergent series, so it's not real old-school math...)
Hint
Integrate term by term to obtain a geometric series (this is allowed inside of the convergence radius of the series).
We can try to find the Taylor Series expansion (http://mathworld.wolfram.com/TaylorSeries.html) about $0$ (also known as a Maclaurin Series (http://mathworld.wolfram.com/MaclaurinSeries.html)) of $(1+x)^{-2}$:
$$\frac{d}{dx} (1+x)^{-2}=-2(1+x)^{-3}\;\;\;\;\;\;\;\;\frac{d^2}{dx^2}(1+x)^{-2}=6(1+x)^{-4}$$ In general: $$\frac{d^n}{dx^n}(1+x)^{-2}=(-1)^n(n+1)!(1+x)^{-(n+2)}$$ Hence $$\frac{d^n}{dx^n}(1+x)^{-2}\Big|_{x=0}=(-1)^n(n+1)!$$ Now: $$(1+x)^{-2}=\sum_{n=0}^{\infty}\frac{d^n}{dx^n}(1+x)^{-2}\Big|_{x=0}\frac{x^n}{n!}=\sum_{n=0}^{\infty}(-1)^n\frac{(n+1)!}{n!}x^n=\sum_{n=0}^{\infty}(-1)^n (n+1)x^n$$ $$=1-2x+3x^2-4x^3+\cdots$$ As we wanted.
$\begin{aligned} \displaystyle \sum_{k=0}^{\infty} (-1)^{k}(k+1)x^{k} & =\sum_{k=0}^{\infty} \sum_{j=0}^{k}(-1)^{k}x^{k} \\& = \sum_{k=0}^{\infty} \sum_{j=0}^{k}(-1)^{k-j}(-1)^jx^{j}x^{k-j} \\& = \bigg(\sum_{k=0}^{\infty}(-1)^kx^k\bigg)\bigg(\sum_{k=0}^{\infty}(-1)^kx^k\bigg) \\& = \bigg(\frac{1}{1+x}\bigg)\bigg(\frac{1}{1+x}\bigg) \\& = \frac{1}{(1+x)^2}.\end{aligned} $
