# An infinite sum

A puzzle from @dmarain:

Explain using at least 2 methods:

— David Marain (@dmarain) August 18, 2016

(1+x²)(1-x²+x⁴-x^6+…)=1, -1<x<1

And why that domain?@johnjoy1966 @aranglancy @HarMath @mrdardy #MTBoS

As usual, have a crack if you want to. Spoilers below the line.

I bet there are dozens of methods here.

### Approach Number 1

My first approach is to just multiply out the brackets: one multiplied by the second bracket is $(1 -x^2 + x^4 - x^6 + \dots)$, and $x^2$ multiplied by the second bracket is $(x^2 - x^4 + x^6 - \dots)$, and all of the terms cancel out except for the 1. We’re allowed to rearrange because the sequence converges absolutely for $|x|<1$.

### Approach Number 2

A cheeky binomial expansion: $(1+x^2)^{-1} = 1 - x^2 + x^4 - x^6 + \dots$ for $|x|<1$, so multiplying this by $(1+x^2)$ gives 1.

### Approach Number 3

The second bracket is a geometric series with a first term of $a=1$ and a common difference of $r= -x^2$. Its sum to infinity is $\frac{a}{1-r}$ (as long as $|r|<1$), which works out to be $\frac{1}{1+x^2}$. Multiplying this by $x^2$ gives 1.

### Approach Number 4

Boringly, divide 1 by $(1+x^2)$ in whatever way you fancy. You’ll get the infinite series in the second bracket.

Part of me says “there has to be a trigonometric way to do this!”, but I can’t immediately see it. There’s probably a calculus approach, too. Any other ideas? I’d love to hear them.