Infinite product $\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$

Question

I have to show that if $|z| < 1$, $z \in \mathbb{C}$, $$\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$$

This exercise if taken from Remmert, Classical Topics in Complex Function Theory.

I don't know where to start. Could anyone give a hint ?

Answer 1

As mr_e_man notes, we have

$$(1+z^n)(1-z^{2n-1})=\frac{(1-z^{2n})(1-z^{2n-1})}{1-z^n}$$

But the numerator covers $1-z^k$ for $k=2n$ and $k=2n-1$, and hence the infinite product is simply one by telescopic cancellation.

This argument relies on the assumption that $\prod_{n=1}^\infty(1-z^n)$ converges, but this can be easily shown by noting that $\sum z^n$ converges.

Answer 2

Let the left be $f(z)$. Do associate law: $(1+z)$ vs $(1-z)$, $(1+z^3)$ vs $(1-z^3)$, $(1+z^5)$ vs $(1-z^5)$, etc., we see $$ f(z)=(1+z^2)(1-z^2)(1+z^4)(1-z^6)(1+z^6)(1-z^{10})...=f(z^2). $$ Now $f(z)=1+c_nz^n+...$, where $c_n$ is the first nonzero coefficient after $c_0=1$, then $1+c_nz^n+...=f(z)=f(z^2)=1+c_nz^{2n}+...$, thus $c_n=0$. So $f=1$.

Answer 3

Hint: Rewrite the product as \begin{eqnarray*} \prod_{k=1}^{\infty} \left( (1-z^{2k-1}) \prod_{j=0}^{\infty} ( 1+z^{(2k-1)2^j}) \right). \end{eqnarray*}

Answer 4

The finite product is $$ \prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right) =\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag1 $$ This can be proven by induction. Suppose $(1)$ is true for $n$ $$ \begin{align} \prod_{k=1}^{n+1}\left(1+z^k\right)\left(1-z^{2k-1}\right) &=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)\tag2\\ &=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag3\\ &=\left(1-z^{2n+2}\right)\left(1-z^{2n+1}\right)\prod_{k=n+2}^{2n}\left(1-z^k\right)\tag4\\ &=\prod_{k=n+2}^{2n+2}\left(1-z^k\right)\tag5 \end{align} $$ Explanation:
$(2)$: pull the $k=n+1$ term out front
$(3)$: apply $(1)$ for $n$
$(4)$: pull the $k=n+1$ term out front
$(5)$: bring the $k=2n+1$ and $k=2n+2$ terms inside

Thus, $(1)$ is true for $n+1$. $\large\square$

Since $$ \begin{align} \left|\sum_{k=n+1}^{2n}\log\left(1-z^k\right)\right| &\le\sum_{k=n+1}^{2n}\left|\log\left(1-z^k\right)\right|\tag7\\ &\le\sum_{k=n+1}^\infty\left|\log\left(1-z^k\right)\right|\tag8\\ &\le\sum_{k=n+1}^\infty\frac{|z|^k}{1-|z|^k}\tag9\\ &\le\frac1{1-|z|^{n+1}}\sum_{k=n+1}^\infty|z|^k\tag{10}\\ &=\frac{|z|^{n+1}}{\left(1-|z|\right)\left(1-|z|^{n+1}\right)}\tag{11} \end{align} $$ Explanation:
$\phantom{1}(7)$: generalized triangle inequality
$\phantom{1}(8)$: absolute value is non-negative
$\phantom{1}(9)$: $|\log(1+z)|\le\frac{|z|}{1-|z|}$
$(10)$: $\frac1{1-|z|^k}\le\frac1{1-|z|^{n+1}}$
$(11)$: sum of a geometric series

Equation $(1)$ and inequality $(11)$ say that $$ \left|\log\left(\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)\right)\right|\le\frac{|z|^{n+1}}{\left(1-|z|\right)\left(1-|z|^{n+1}\right)}\tag{12} $$ Thus, $$ \log\left(\prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)\right)=0\tag{13} $$ and therefore, $$ \prod_{k=1}^\infty\left(1+z^k\right)\left(1-z^{2k-1}\right)=1\tag{14} $$