1. infinite series problem?

(1+x^2)(1+x^4)(1+x^8)...(goes on forever)

|x|< 1

all i am told to do is simplify and i really have no idea where to start on this. please help.

2. nm, below

3. Hello, oblixps!

Simplify: . $S \:=\:(1+x^2)(1+x^4)(1+x^8)(1 + x^{16}) \hdots\qquad |x|\,<\, 1$

We have: . $S \;=\;(1+x^2)(1+x^4)(1+x^8)(1+x^{16}) \hdots$

Multiply both sides by $(1-x^2)$

. $(1-x^2)S \;=\;\underbrace{(1-x^2)(1+x^2)}(1+x^4)(1+x^8)(1+x^{16}) \hdots$

. . . . . . . . . . $= \;\underbrace{(1-x^4)(1+x^4)}(1+x^8)(1+x^{16}) \hdots$

. . . . . . . . . . . . $= \;\underbrace{(1 - x^8)(1 + x^8)}(1 + x^{16}) \hdots$

. . . . . . . . . . . . . . . $= \;(1 - x^{16})(1 + x^{16})\hdots$

The right side is of the form: . $\left(1 - x^{2^n}\right)$

Since $|x|\,<\,1$, then: . $\lim_{n\to\infty}\left(1 - x^{2^n}\right) \;=\;1 - 0 \;=\;1$

Our equation becomes: . $(1-x^2)S \;=\;1 \quad\Rightarrow\quad\boxed{ S \;=\;\frac{1}{1-x^2}}$

4. Let's be brave:

$f(x) = {\color{blue}\left(1+x\right)\left(1+x^2\right)}\l eft(1+x^4\right)\left(1+x^8\right)\left(1+x^{16}\r ight)\cdots$

$= {\color{blue}\left(1 +x + x^2 + x^3 \right)}\left(1+x^4\right)\left(1+x^8\right)\left( 1+x^{16}\right)\cdots$

$= \left(1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\right)\left(1+x^8\right)\left(1 + x^{16}\right)\cdots$

see the pattern?

$= \sum_{ n = 0 }^{\infty} x^n$

which is the infinite series representation of the known function: $f(x) = \frac{1}{1-x}$

5. oh wow thanks!! i was kinda weirded out by the going on forever part that i didn't even think about multiplying the first few terms together. this problem makes complete sense now! thanks a bunch guys!