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I want to prove that $\prod_{n=0}^\infty(1+z^{2^n})=(1-z)^{-1}$ for all $|z|<1$.

By multiplying $1-z$ to both sides, the equation becomes $$(1-z)(1+z)(1+z^2)(1+z^4)\ldots=1$$

Multiplying the first pair on the left yields $$(1-z^2)(1+z^2)(1+z^4)\ldots=1$$

And then

$$(1-z^4)(1+z^4)\ldots=1$$

And $|z|^{2^n}$ keeps getting smaller and smaller since $|z|<1$.

But the product is infinite, so this doesn't seem very rigorous? How can I make it rigorous?

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    $\begingroup$ hint: What you want is $$\lim_{k\to\infty}\prod_{n=0}^k(1+z^{2^n})=(1-z)^{-1}$$. Can you show $$\prod_{n=0}^k(1+z^{2^n})=\frac{1-z^{2^{k+1}}}{1-z}$$? $\endgroup$ – Apple Nov 20 '13 at 20:12
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Any integer $j$ with $0\leq j\leq2^n-1$ has a unique binary representation of the form $$j=\sum_{k=0}^{n-1} b_k\>2^k,\qquad b_k\in\{0,1\}\ .$$ Therefore these $j$ are in bijective correspondence with the $2^n$ subsets of the set $\{1,2,4,8,\ldots,2^{n-1}\}$.

When the product $$P_n:=\prod_{k=0}^{n-1}\left(1+z^{2^k}\right)=(1+z)(1+z^2)(1+z^4)\cdot\ldots\cdot(1+z^{2^{n-1}})$$ is expanded one obtains the corresponding term $z^j$ for each of these subsets. It follows that for $|z|<1$ one has $$P_n=\sum_{j=0}^{2^n-1} z^j={1-z^{2^n}\over 1-z}\to{1\over1-z}\qquad(n\to\infty)\ .$$

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The exponent is $2^k$ at each step. As k tends towards infinity, so does $2^k$, meaning that $z^{2^k}\to0$, since $|z|<1$. And $1-0=1$. QED.

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  • $\begingroup$ I agree that $z^{2^k}\rightarrow 0$, but in the $k$-th step there are still terms $(1+z^{2^k})(1+z^{2^{k+1}})\ldots$. How can you take care of those? $\endgroup$ – JJ Beck Nov 20 '13 at 20:27
  • $\begingroup$ Are you familiar with the mathematical concepts of limits, infinite products, infinite series, and their computation ? $\endgroup$ – Lucian Nov 20 '13 at 20:49

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