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Prove the following equation of complex power series.

I must show for $|z|<1$ that $$\sum_{k=0}^\infty \frac{z^{2^k}}{1-z^{2^{k+1}}} = \frac{z}{1-z}.$$ I'm not really sure where to start or how to simplify the problem at all. Any help would be great.

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marked as duplicate by Micah, Cameron Buie, Calvin Lin, David Moews, Thomas Feb 5 '13 at 1:09

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What do you get for the partial sum $S_n$? Note that $(1-z^{2^{k+1}})=(1-z^{2^k})(1+z^{2^k})$. –  Calvin Lin Feb 5 '13 at 0:37

1 Answer 1

This looks nasty, but with the following approach it should boil down to a straight forward calculation:

You can write the RHS as a power-series: $$\frac{z}{1-z} = z \sum_{k=0}^{\infty} z^k = \sum_{k=0}^{\infty} z^{k+1}$$

Now try to rewrite you LHS as power-series as well and compare the coeffecients: $$\sum_{k=0}^{\infty} \frac{z^{2^k}}{1-z^{2^{k+1}}} = \sum_{k=0}^{\infty} \left(z^{2^k} \frac{1}{1 - z^{2^{k+1}}}\right) = \sum_{k=0}^{\infty} \left(z^{2^k} \sum_{j=0}^{\infty} \left(z^{2^{k+1}}\right)^j\right) \\= \sum_{k=0}^{\infty} \left(z^{2^k} \sum_{j=0}^{\infty} z^{2^{k+1}j}\right) = \sum_{k=0}^{\infty}\sum_{j=0}^{\infty} z^{2^k + j2^{k+1}}$$

To see that both power-series are the same note that every $n \in \mathbb{Z}$, $n\geq 1$ there exists unique integers $k$ and $j$ such that $n=2^k + j2^{k+1}$ (namely $2^k$ is the gratest power of $2$ which divides $n$).

I guess there are more elegant ways to prove this, but since you didn't know how to start it might be usefull to see this approach.

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How exactly does the LHS break down like that? That is the only part I can't follow. The step with the double summation throws me off –  Moses G Feb 5 '13 at 4:26
    
I added two more steps to the calculation. I use the geometric series to rewrite the fraction as a power-series: $1/(1-w) = 1 + w^2 + w^3 + \cdots$ with $w = z^{2^{k+1}}$. –  Sam Feb 5 '13 at 9:43

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