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This is exercise 21 of Chapter 1 from Stein and Shakarchi's Complex Analysis.

Show that for $|z|<1$ one has $$\frac{z}{1-z^2}+\frac{z^2}{1-z^4}+\cdots +\frac{z^{2^n}}{1-z^{2^{n+1}}}+\cdots =\frac{z}{1-z}$$and

$$\frac{z}{1+z}+\frac{2z^2}{1+z^2}+\cdots \frac{2^k z^{2^k}}{1+z^{2^k}}+\cdots =\frac{z}{1-z}.$$

Justify any change in the order of summation.

[Hint: Use the dyadic expansion of an integer and the fact that $2^{k+1}-1=1+2+2^2+\cdots +2^k$.]

I don't really know how to work this through. I know that $\frac{z}{1-z}=\sum_{n=1}^\infty z^n$ and each $n$ can be represented as a dyadic expansion, but I don't know how to progress from here. Any hints solutions or suggestions would be appreciated.

nomadicmathematician
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  • Likely, the answer lies in the partial sums and some creative multiplication. – Simply Beautiful Art Mar 04 '17 at 01:23
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    For the first one, you can expand each of the individual terms into using their geometric series expansions. Then you'd have to show they exhaust (and don't repeat) all the whole numbers. The power seem to be: $2k + 1$, $4k + 2$, ..., $2^{n+1} k + 2^n$ for $k = 0,1,2...$. And the second one likely works in a similar fashion except with alternating sums... I'm not sure how to write this out properly. – minimalrho Mar 04 '17 at 02:26
  • There is a similar question http://math.stackexchange.com/questions/1628569/prove-that-sum-n-0-infty-fracz2n1-z2n1-fracz1-z with no accepted answer though ... – rtybase Mar 04 '17 at 12:05

3 Answers3

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Since minimalrho has explained how to proceed with the given hint, I'll give an alternative method. The $k$th summand of the first series can be written

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

and the $k$th summand of the second series can be written

$$\frac{2^kz^{2^k}}{1 - z^{2^k}} - \frac{2^{k+1}z^{2^{k+1}}}{1-z^{2^{k+1}}}$$

Hence, the $N$th partial sums of the two series telescope to

$$\frac{z}{1 - z} - \frac{z^{2^{N+1}}}{1 - z^{2^{N+1}}}\quad \text{and}\quad \frac{z}{1 - z} - \frac{2^{N+1}z^{2^{N+1}}}{1 - z^{2^{N+1}}}$$

respectively. Using the condition $\lvert z\rvert < 1$, argue that $z^{2^{N+1}}/(1 - z^{2^{N+1}})$ and $2^{N+1}z^{2^{N+1}}/(1 - z^{2^{N+1}})$ tend to $0$ as $N\to \infty$. Then the results follow.

kobe
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Hint and partial answer: Using this partition of integers and $$\frac{z}{1-z}=\sum_{n=1}z^n=...$$ This series is absolute converging given $|z|<1$, thus changing the order of summation doesn't affect the final value. As a result: $$...=\sum_{k=0}\left(\sum_{t\in A_k}z^t\right)=\sum_{t\in A_0}z^t+ \sum_{k=1}\left(\sum_{t\in A_k}z^t\right)=\sum_{s=0}z^{2s+1} + \sum_{k=1}\left(\sum_{s=0}z^{2^k(2s+1)}\right)=\\ z\sum_{s=0}z^{2s}+\sum_{k=1}z^{2^k}\left(\sum_{s=0}z^{2^k(2s)}\right)=\frac{z}{1-z^2}+\sum_{k=1}\frac{z^{2^k}}{1-z^{2^{k+1}}}$$

Community
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rtybase
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Hint for the first sum. Note that each positive integer $n$ can be written in a unique way as the product of a power of $2$, $2^k$, and an odd number $(2j+1)$.

Hint for the second sum. Note that if $n=2^k(2j+1)$ then the coefficient of $z^n$ of the left-hand side is $$-1-2-2^2-\cdots -2^{k-1}+2^{k}.$$

Robert Z
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