$(3x- 3)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{2^{-n}}+1}=(x^3-1)\prod_{n = 1}^∞\frac{\exp(2^{-n})+1}{x^{3 \cdot 2^{-n}}+1}$

Question

Show that

$$ (3x - 3) \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {x^{2^{-n}} + 1} = (x^3 - 1) \prod_{n = 1}^{\infty} \frac{ \exp(2^{-n}) +1} {x^{3\cdot2^{-n}} + 1} $$

Answer 1

Hint: $\;\; \left(x^{2^{-n}}-1\right) \cdot \left(x^{2^{-n}}+1\right)\left(x^{2^{-(n-1)}}+1\right)\ldots\left(x^{2^{-1}}+1\right) = x - 1 \,$ by telescoping.

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[ EDIT ]  It then follows that the $\,n^{th}\,$ partial product on the LHS is:

$$\require{cancel} 3\cancel{(x - 1)} \cdot \frac{\left(x^{2^{-n}} - 1\right)}{\left(e^{2^{-n}} - 1\right)} \cdot \frac{\left(e^{2^{-n}} - 1\right) \prod_{k = 1}^{n} \left(e^{2^{-k}} +1\right)}{\cancel{\left(x^{2^{-n}} - 1\right)\prod_{k = 1}^{n} \left(x^{2^{-k}} +1\right)}} = 3(e-1) \cdot \frac{x^{2^{-n}} - 1}{e^{2^{-n}} - 1} $$

For $\,n \to \infty\,$ $\,\,2^{-n} \to 0\,$, and the LHS therefore reduces to:

$$ \lim_{a \to 0} \,3(e-1) \cdot \frac{x^a - 1}{e^{a} - 1} = \ldots $$