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Is there a way to rewrite a series like $\sum_{n=1}^\infty \frac{z^n}{1-z^n}$ as a power series? I know it's holomorphic on the unit disk, so I'm curious if there is some way to do so.

3 Answers 3

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Since $\frac{z^n}{1-z^n}=\sum_{\{m\in\mathbb{N}_0:n|m\}}z^m$ you have: $\sum_{n=1}^{+\infty}\frac{z^n}{1-z^n}=\sum_{m=1}^{+\infty} d(m)\,z^m.$

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If $\,|z|<1\,$ , then

$\frac{1}{1-z^n}=1+z^n+z^{2n}+\cdots\Longrightarrow \frac{z^n}{1-z^n}=z^n+z^{2n}+z^{3n}+\cdots$

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We have

$\frac{1}{1 - z} = \sum_{k=0}^{\infty} z^k$.

So,

$\frac{1}{1 - z^n} = \sum_{k=0}^{\infty} z^{kn}$.

Then,

$\frac{z^n}{1 - z^n} = \sum_{k=0}^{\infty} z^{(k+1)n}$.

Now,

$\sum_{n=1}^{\infty} \frac{z^n}{1 - z^n} = \sum_{n=1}^{\infty} \sum_{k=0}^{\infty} z^{(k+1)n} = \sum_{n=1}^{\infty} \sigma_0(n) z^n$

where $\sigma_0(n)$ is the number of divisors of $n$, or, equivalently, the number of ways to write n as a product of two positive integers, order not mattering. Note that the number of times $(j+1)k$, $j \ge 0$, $k \ge 1$ equals $n$ is given by this amount.