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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?

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    $\begingroup$ Just to say, this is not a sequence, but a series. $\endgroup$ – ajotatxe Feb 19 '15 at 23:29
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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 $$ f(x)=\frac1{(1+x)^2} $$

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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.

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Hint

Integrate term by term to obtain a geometric series (this is allowed inside of the convergence radius of the series).

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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...)

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    $\begingroup$ What an extremely ironical use of the term "blindly manipulating infinite series". I'm not sure if you're aware, but Euler was actually almost completely blind for the last 17 years of his life. $\endgroup$ – Uncountable Feb 20 '15 at 0:13
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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.

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$\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} $

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