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Show that for $|z| \lt 1$ with $z \in \Bbb C$, we have

$$ \sum_0^\infty \frac{{z^2}^k}{1-{z^2}^{k+1}} = \frac{z}{1-z} $$

$$ \sum_0^\infty \frac{2^k{z^2}^k}{1+{z^2}^{k}} = \frac{z}{1-z} $$

My guess is that the second one is obtained by derivating the first one or something like that, but I can't manage to prove the first one.

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Stop doing stupid edits! –  user26857 Feb 4 '13 at 23:56
    
What YACP means is: stop vandalizing the post. –  Cameron Buie Feb 5 '13 at 0:47

1 Answer 1

up vote 7 down vote accepted

We have $$\sum_{k=0}^{\infty}\dfrac{z^{2^k}}{1-z^{2^{k+1}}} = \sum_{k=0}^{\infty}\sum_{l=0}^{\infty} z^{2^k + l \cdot 2^{k+1}} = \sum_{l=0}^{\infty} \sum_{k=0}^{\infty}z^{2^k(2l+1)} = \sum_{m=1}^{\infty} z^m$$ since there is a unique decomposition of any number $m \in \mathbb{Z}^+$ as $2^k(2l+1)$. But $$\sum_{m=1}^{\infty} z^m = \dfrac{z}{1-z}$$ Hence, we get that $$\sum_{k=0}^{\infty}\dfrac{z^{2^k}}{1-z^{2^{k+1}}} = \dfrac{z}{1-z}$$

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