Consider:
$$\sum_{k=0}^{\infty} \frac{x^{2^k}}{\prod_{i = 0}^k (x^{2^i}+1)}$$ For $x>1$.
The first term is $\frac{x}{x+1}$ which can be expanded as $x-x^2+ \cdots$. Every term after the initial of the initial sum has at least $x^2$ in the numerator so it follows that the big sum should have a $x$ term when simplified.
According to $pg. 170$ ($180$ in the file) of the book here: https://cambomaths.files.wordpress.com/2014/02/red-book-of-mathematical-problems-the-dover-books-on-mathematics-williams-hardy-2010-0486694151-185p.pdf
We have the sum is actually $1$. What am I doing wrong?
\begin{align} \sum_{k=0}^n\frac{x^{2^k}}{\prod_{i = 0}^k (x^{2^i}+1)} &=(x-1)\sum_{k=0}^n\frac{x^{2^k}}{(x-1)\prod_{i = 0}^k (x^{2^i}+1)}\\ &=(x-1)\sum_{k=0}^n\frac{x^{2^k}}{x^{2^{k+1}}-1}\\ &=(x-1)\sum_{k=0}^n\left(\frac{x^{2^k}+1}{x^{2^{k+1}}-1}-\frac 1{x^{2^{k+1}}-1}\right)\\ &=(x-1)\sum_{k=0}^n\left(\frac{1}{x^{2^k}-1}-\frac 1{x^{2^{k+1}}-1}\right)\\ &=(x-1)\left(\frac{1}{x-1}-\frac 1{x^{2^{n+1}}-1}\right)\\ &\xrightarrow{n\to\infty}1 \end{align}