$\sum_{n=0}^\infty z^n = \prod_{m=0}^\infty \left(1+z^{2^m}\right)$

Question

When reading Iwaniec and Kowalski's Analytic Number Theory, I came across the following "identity" on page 11 (the Amazon link has a free book preview which includes page 11):

$$\sum_{n=0}^\infty z^n = \prod_{m=1}^\infty \left(1+z^{2^m}\right).$$

However, I don't think this is true, as there will never be odd exponent terms in the product expansion. I think the identity should, instead, be $$\sum_{n=0}^\infty z^n = \prod_{m=0}^\infty \left(1+z^{2^m}\right).$$

Here's what I believe to be a proof of the equality between the sum and product.

Let $s_N=\sum_{n=0}^Nz^n$ and $p_N=\prod_{n=0}^N\left(1+z^{2^n}\right)$. I claim, first, that $p_N=s_{2^{N+1}-1}$ for all $N\geq0$, which I will prove by induction.

For $N=0$, we have $p_0=1+z=s_1=s_{2^{0+1}-1}$. Now, suppose that this is true for $N$, we will show that this implies the same for $N+1$, that is, $p_{N+1}=s_{2^{N+2}-1}$.

If $p_N=s_{2^{N+1}-1}$, then $$p_{N+1}=s_{2^{N+1}-1}\left(1+z^{2^{N+1}}\right)=\left(1+z+\cdots+z^{2^{N+1}-1}\right)\left(1+z^{2^{N+1}}\right)=\\\left(1+z+\cdots+z^{2^{N+1}-1}\right)\cdot1+\left(1+z+\cdots+z^{2^{N+1}-1}\right)\cdot z^{2^{N+1}}=\\1+z+\cdots+z^{2^{N+1}-1}+z^{2^{N+1}}+\cdots+z^{2^{N+1}+2^{N+1}-1}=\\1+z+\cdots+z^{2^{N+1}-1}+z^{2^{N+1}}+\cdots+z^{2^{N+2}-1}=s_{2^{N+2}-1},$$ as required.

Therefore, since $|p_N-s_{2^{N+1}-1}|=0<\varepsilon$ for any $\varepsilon>0$ and all $N\geq0$, the identity is proven. That is, the sum converges for a particular $z$ if and only if the product also converges for that $z$, and, in that case, they both converge to the same value.

Am I correct that the identity in Iwaniec/Kowalski wrong, and, if so, does my attempted proof of a corrected version work?

Answer 1

You are right, the product needs to start at $m = 0$. I think it's a plain typo in the book.

Your proof is correct, but at the end you seem to be a little afraid of its simplicity, after you have shown the identity $p_N = s_{2^{N+1}-1}$, you are effectively done. Since the series converges for $\lvert z\rvert < 1$, and the sequence of partial products is a subsequence of the sequence of partial sums, it follows that the identity holds for all $z$ with $\lvert z\rvert < 1$. Since the factors of the product converge to $1$ only when $\lvert z\rvert < 1$, the product does not converge for any $z$ for which the sum doesn't converge.

Answer 2

Another proof of $$\prod_{n\ge0}(1+z^{2^n})=\sum_{n\ge0}z^n=\frac1{1-z}$$ is as follows. Define the pair of functions $f,g$ as $$\begin{align} f(z)&=\prod_{n\ge0}(1-z^{2^n})\\ g(z)&=\prod_{n\ge0}(1+z^{2^n}), \end{align}$$ for $|z|<1$. Then we see that $$\begin{align} f(z)g(z)&=\prod_{n\ge0}(1-z^{2^n})(1+z^{2^n})\\ &=\prod_{n\ge0}(1-z^{2\cdot2^n})\tag 1\\ &=\prod_{n\ge0}(1-(z^2)^{2^n})\\ &=f(z^2). \end{align}$$ However, note that $$f(z^2)=\prod_{n\ge0}(1-z^{2\cdot2^n})=\prod_{n\ge0}(1-z^{2^{n+1}})=\prod_{n\ge1}(1-z^{2^n})=\frac{1}{1-z}\prod_{n\ge0}(1-z^{2^n})=\frac{f(z)}{1-z}.$$ Thus, from $(1)$, $$f(z)g(z)=\frac{f(z)}{1-z}.$$ That is, $$g(z)=\prod_{n\ge0}(1+z^{2^n})=\frac{1}{1-z}.$$