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Let $f(z)=\sum\limits_{k=1}^\infty\frac{z^k}{1-z^k}$. I want to show that this series represents a holomorphic function in the unit disk. I'm, however, quite confused. For example, is $f(z)$ even a power series? It doesn't look as such. Here's what I have so far come up with.

Proof:

$$ \sum\limits_{k=1}^\infty\frac{z^k}{1-z^k}=-\sum\limits_{k=1}^\infty\frac{1}{1-z^{-k}}=-\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty z^{kn}$$

[If the above expression is correct then we have a Laurent series of a power series].

Now, $|c_n|=1$ for all $n$. By Parseval's formula,

$$2\pi\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty \rho^{2kn}=\sum\limits_{k=-\infty}^{-1} \int\limits_0^{2\pi}\left|g(\rho^ke^{it})\right|^2dt=^? \int\limits_0^{2\pi} \sum\limits_{k=1}^{\infty} \left|g(\rho^{-k}e^{it})\right|^2dt$$ where $0\le\rho\le 1$, and $g$ is some function (holomorphic) on the unit disk. So we know that this integral (above in the middle) exists.

Now, I believe, what remains to be proved is that the infinite series of this integral also exists. Do you think this approach is OK, or did I make some mistakes in it? How can we prove that the series $\sum\limits_{k=-\infty}^{-1}\sum\limits_{n=0}^\infty \rho^{2kn}$ converges? And then, since it converges, does this imply that $G$ is holomorphic?

Thanks a lot.

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    $\begingroup$ You're right, it's not a power series. $\endgroup$ – zhw. Mar 6 '17 at 18:59
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Suppose $0<r<1.$ Then $r^k \to 0.$ Thus $r^k\le 1/2$ for large $k.$ For these $k$ and $|z|\le r$ we have

$$\left| \frac{z^k}{1-z^k}\right | \le \frac{r^k}{1-r^k} \le 2r^k.$$

Since $\sum 2r^k<\infty,$ Weierstrass M implies our series converges uniformly on $\{|z|\le r\}.$ This is true for every $r\in (0,1),$ hence the series converges uniformly on compact subsets of the open unit disc. This proves the series converges to a homolorphic function in the open unit disc.

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Hint: Observe \begin{align} f_k(z) = \frac{z^k}{1-z^k} \end{align} is definitely holomorphic on the open unit disc for all $k$. In particular, we have that \begin{align} g_n(z) = \sum^n_{k=1}f_k(z) \end{align} is also holomorphic on the unit disc. Now, show that $g_n$ is a normally convergent sequence, i.e. show that $g_n$ converges uniformly on compact subsets of the unit disc.

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Theorem : if $\sum_{n=0}^\infty a_n z^n$ converges absolutely for $|z| < R$ then it is analytic/holomorphic for $|z| < R$.

For $|z| < 1$ : $$\sum_{k=1}^\infty\frac{z^k}{1-z^k}=\sum_{k=1}^\infty\sum_{n=1}^\infty z^{kn} = \sum_{m=1}^\infty z^{m} \tau(m)$$ where $\tau(m)=\sum_{d | m} 1$ is the number of divisors. Since $\tau(m) <m$ it means that $\sum_{m=1}^\infty z^{m} \tau(m)$ converges absolutely for $|z| < 1$. Hence changing the order of summation doesn't matter, and $\sum_{m=1}^\infty z^{m} \tau(m)=\sum_{k=1}^\infty\frac{z^k}{1-z^k}$ is analytic/holomorphic on $|z| < 1$.

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  • $\begingroup$ Why is your summation in the middle series beginning from $k=1$ to $\infty$, and not $-\infty$ to $-1$? WE don't seem to be accounting for the negative powers of $z$. $\endgroup$ – sequence Mar 6 '17 at 5:43
  • $\begingroup$ @sequence because I corrected your mistake. for $|z| < 1$ : $\frac{1}{1-z} = \sum_{n=0}^\infty z^n$ but for $|z| > 1$ :$\frac{1}{1-z} =\frac{-z^{-1}}{1-z^{-1}} =-\sum_{n=1}^\infty z^{-n}$ $\endgroup$ – reuns Mar 6 '17 at 5:58
  • $\begingroup$ Sorry, I'm somewhat puzzled, I don't see what exactly my mistake is. $\endgroup$ – sequence Mar 6 '17 at 6:03
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    $\begingroup$ @sequence what you wrote ($\sum_{n=1}^\infty z^{-n}$) doesn't converge for $|z| \le 1$ $\endgroup$ – reuns Mar 6 '17 at 6:04
  • $\begingroup$ I see. Can you please show how exactly you derived $$\sum_{k=1}^\infty\frac{z^k}{1-z^k}=\sum_{k=1}^\infty\sum_{n=1}^\infty z^{kn} $$? Also, what does $\tau$ mean? The number of divisors of what? @user1952009 $\endgroup$ – sequence Mar 6 '17 at 6:43

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