How to evaluate the following limit $\lim_{n\to \infty} \prod_{k=0}^n \left(1+\delta^{2^{k}}\right)$?

Question

If $\delta \in ]0,1[$ then the following limit is : $$\lim_{n\to \infty} \prod_{k=0}^n \left(1+\delta^{2^{k}}\right)$$ I tried to simplify the following product : $$\begin{align} \prod_{k=0}^n \left(1+\delta^{2^{k}}\right)&= (1+\delta)(1+\delta^2)...\left(1+\delta^{2^{n}}\right)\\ \ln\left(\prod_{k=0}^n \left(1+\delta^{2^{k}}\right)\right) &=\sum_{k=0}^n \ln\left(1+\delta^{2^{k}}\right) \end{align}$$ I couldn't really go far from here I tried the following : $$0<\delta^{2^{k}}<1 \Leftrightarrow 0<\ln\left(1+\delta^{2^{k}}\right)<\ln(2)$$ But it seems useless.

Any idea to evaluate the product hence the limit ?

Answer 1

Because it seems like you're asking for a hint more than a full answer, that's what I'll go for. Consider the first few partial products:

$$1 + \delta$$

$$(1 + \delta)(1 + \delta^2) = 1 + \delta + \delta^2 + \delta^3$$

$$(1 + \delta)(1 + \delta^2)(1 + \delta^4) = 1 + \delta + \delta^2 + \delta^3 + \delta^4 + \delta^5 + \delta^6 + \delta^7$$

I don't know about you, but I see a pretty striking pattern going on here, relating the infinite product to something that's much more familiar to me. The only thing missing is a proof that the pattern I'm seeing continues, which can be done using induction.

Answer 2

$$\lim_{n\to \infty}\prod_{k=1}^n\frac{(1-\delta)(1+\delta^{2^k})}{1-\delta}=\lim_{n\to \infty}\frac{1-\delta^{2^{n+1}}}{1-\delta}=\frac{1}{1-\delta}$$