Demonstrate recursively that $\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$

Question

Demonstrate recursively that

$$\prod_{k = 0}^\infty (1 + x^{2^k}) = \frac{1}{1-x}$$

My work:

Define

$$a_n = \prod_{k = 0}^n (1 + x^{2^k}) = (1 + x^{2^n})a_{n - 1} \iff a_n - (1 + x^{2^n})a_{n - 1} = 0$$ $$A(x) = \sum_{n = 0}^\infty a_nx^n = a_0 + \sum_{n = 1}^\infty a_{n}x^n = 1 + x + \sum_{n = 1}^\infty a_{n}x^n$$ $$\implies xA(x) = \sum_{n = 1}^\infty a_{n-1}x^n$$

Therefore,

\begin{align} A(x) - (1 + x^{2^n})xA(x) &= 1 + x + \sum_{n = 1}^\infty a_{n}x^n - \sum_{n = 1}^\infty (1 + x^{2^n})a_{n-1}x^n\\ A(x) \left(1 - x(1 + x^{2^n})\right) &=1 + x + \sum_{n = 1}^{\infty} (a_n - (1 + x^{2^k})a_{n - 1})x^n\\ A(x) \left(1 - x(1 + x^{2^n})\right) &=1 + x\\ A(x) &= \frac{1 + x}{1 - x(1 + x^{2^n})}\\ &= \frac{1 + x}{1 - x - x^{2^n + 1}} \end{align}

To find $a_n$, I now want to transform $A(x)$ in the form $\sum_{k = 0}^\infty a_nx^n$ and then take the limit as $n \to \infty$. However, I’m unsure how to do this since I don’t think the expression can be decomposed into partial fractions.

My question:

• Is my approach correct?
• How do I continue?

Note: I know that there are other solutions, but I specifically want to see if defining a recurrence can yield a solution.

Answer 1

(This may be a duplicate answer, but I worked it out myself, so, whatever.)

As in other answer answers, let $a_n = \prod_{k = 0}^n (1 + x^{2^k}) $ and $b_n=(1-x)a_n$.

Then, if $n \ge 2$,

$\begin{array}\\ b_n &= (1-x)\prod_{k = 0}^n (1 + x^{2^k})\\ &= (1-x)(1+x)\prod_{k = 1}^n (1 + x^{2^k})\\ &= (1-x^2)\prod_{k = 1}^n (1 + x^{2^k})\\ &= (1-x^2)(1+x^2)\prod_{k = 2}^n (1 + x^{2^k})\\ &= (1-x^4)\prod_{k = 2}^n (1 + x^{2^k})\\ \end{array} $

I claim that, for $m \le n$, $b_n = (1-x^{2^m})\prod_{k = m}^n (1 + x^{2^k}) $.

Proof.

This is true for $m=0, 1$.

If true for $m<n$ then

$\begin{array}\\ b_n &= (1-x^{2^m})\prod_{k = m}^n (1 + x^{2^k})\\ &= (1-x^{2^m})(1+x^{2^m})\prod_{k = m+1}^n (1 + x^{2^k})\\ &= (1-x^{2^{m+1}})\prod_{k = m+1}^n (1 + x^{2^k})\\ \end{array} $

Setting $m=n-1$,

$\begin{array}\\ b_n &= (1-x^{2^n})\prod_{k = n}^n (1 + x^{2^k})\\ &= (1-x^{2^n})(1 + x^{2^n})\\ &= (1-x^{2^{n+1}})\\ \end{array} $

Therefore, if $|x| < 1$, $\lim_{n \to \infty} b_n =1 $ so $\lim_{n \to \infty} a_n =\dfrac1{1-x} $.

Answer 2

Let's say that $A_n=\prod_{k=0}^n(1+x^{2^k})$. We have $A_0=1+x$. We will prove that $B_n=\sum_{k=0}^{2^n-1}x^k$ is equal to $A_n$. It is clear that $B_0=1+x=A_0$.

So we assume that $A_i=B_i$ and will prove that this implies $B_{i+1}=A_{i+1}$.

\begin{align} A_{i+1}&=\prod_{k=0}^{i+1}(1+x^{2^k})\\ &=\left(1+x^{2^i}\right)\prod_{k=0}^{i}(1+x^{2^k})\\ &=A_i+x^{2^i}A_i\\ &=B_i+x^{2^i}B_i\\ &=\sum_{k=0}^{2^i-1}x^k+x^{2^i}\sum_{k=0}^{2^i-1}x^k\\ &=\sum_{k=0}^{2^{i+1}-1}x^k\\ &=B_{i+1} \end{align}