Show that $\frac{z}{1-z} = \sum_{j=0}^∞ \frac{2^j}{1 + z^{-2^j}}$ when $z ∈ \mathbb{D}$

Question

The question

Knowing that with $z ∈ \mathbb{D}$: $$ \prod_{k=0}^∞(1 + z^{2^k}) = \frac{1}{1-z} $$

prove that with $z ∈ \mathbb{D}$: $$ \sum_{j = 0}^∞ \frac{2^j}{1 + z^{-2^j}} = \frac{z}{1-z} $$

What I've tried

$$ \sum_{j = 0}^∞ \frac{2^j}{1 + z^{-2^j}} = \sum_{j=0}^∞ \frac{2^j z^{2^j}}{1+z^{2^j}} = \frac{1}{1+z} + \frac{2z^2}{1+z^2} + \frac{4z^4}{1+z^4} + … =$$ $$ = \frac{\prod_{k=1}^∞(1+z^{2^k}) + 2z^2 \prod_{k≠1}^∞(1+z^{2^k}) + 4z^4 \prod_{k≠2}^∞(1+z^{2^k})}{\prod_{k=0}^∞ (1 + z^{2^k})} = $$ $$ = (1 - z) \left[ \prod_{k=1}^∞(1+z^{2^k}) + 2z^2 \prod_{k≠1}^∞(1+z^{2^k}) + 4z^4 \prod_{k≠2}^∞(1+z^{2^k}) \right]$$

But I don't know what else I can do right know. Any idea?

Answer 1

Taking the "logarithmic derivative" $f'/f = (\log f)'$ on both sides of the equation $$ \prod_{k=0}^\infty (1 + z^{2^k}) = \frac{1}{1-z} $$ gives $$ \sum_{k=0}^\infty \frac{2^k z^{2^k-1}}{1 + z^{2^k}} = \frac{1}{1-z} $$ which is the desired result.

See $\frac{f'(z)}{f(z)}= \sum_{n=1}^{+ \infty}\frac{f'_{n}(z)}{f_{n}(z)}$ for a justification why this is allowed with an infinite product.