Find the natural boundary of $\sum_{n=1}^\infty \frac{z^n}{1-z^n}$ I'm asked to prove that the natural boundary of $\sum_{n=1}^\infty \frac{z^n}{1-z^n}$ is the unit circle.  
My try: First, use root test to show that the series converges for $|z|<1$. Then I have to show that every point on the unit circle is singular, or equivalently, there is a dense set of singular point on the unit circle. However, I didn't succeed.  
Can anyone give a hint? 
Note: The series can be rewritten as $\sum_{n=1}^\infty d(n)z^n$, where $d(n)$ is the number of positive divisors of $n$. (Maybe it is helpless.)
 A: Let $z_{r}=re^{2\pi ip/q}$, where $p$ and $q>1$ are coprime integers and $0<r<1$. We will show that
\begin{align*}
\lim_{r\rightarrow 1^{-}}\left(1-r\right)\left|f(z_{r})\right|=+\infty, \tag{1}
\end{align*}
whence $f$ admits no analytic continuation $g$ in a neighborhood of $z=e^{2\pi i p/q}$, since that would imply
\begin{align*}
\lim_{r\rightarrow 1^{-}}\left(1-r\right)\left|f(z_{r})\right|=\lim_{r\rightarrow 1^{-}}\left(1-r\right)\left|g(z_{r})\right|=0,
\end{align*}
by the continuity of $g$ at $z$. That the unit circle is the natural boundary of $f$ then follows, since the roots of unity are a dense subset of the unit circle, and the set of singular points is closed.
Decompose the series giving $f(z)$ as
\begin{align*}
f(z)=\sum_{{n\geq 1}\atop{n\equiv 0\pmod{q}}}\dfrac{z^{n}}{1-z^{n}}+\sum_{{n\geq 1}\atop {n\not\equiv 0\pmod{q}}}\dfrac{z^{n}}{1-z^{n}}, \qquad\left|z\right|<1 \tag{2}
\end{align*}
We estimate the modulus of the first series first. Observe that
\begin{align*}
\left|\sum_{{n\geq 1}\atop{n\equiv 0\pmod{q}}}\dfrac{z^{n}}{1-z^{n}}\right|=\sum_{m=1}^{\infty}\dfrac{r^{qm}}{1-r^{qm}},
\end{align*}
and
\begin{align*}
\left(1-r\right)\sum_{m=1}^{\infty}\dfrac{r^{qm}}{1-r^{qm}}&=\dfrac{1-r}{1-r^{q}}\sum_{m=1}^{\infty}\dfrac{1-r^{q}}{1-r^{qm}} r^{mq}\\
&=\dfrac{1}{1+r+\cdots+r^{q-1}}\sum_{m=1}^{\infty}\dfrac{r^{mq}}{1+r^{q}+\cdots+r^{(m-1)q}} \tag{3}\\
&\geq\dfrac{1}{q}\sum_{m=1}^{\infty}\dfrac{r^{mq}}{m}\\
&=\dfrac{-\ln\left|1-r^{q}\right|}{q}=q^{-1}\ln\left|\dfrac{1}{1-r^{q}}\right| \tag{4}
\end{align*}
where we use the geometric series formula in (2) and the Taylor expansion of $\ln\left|1+z\right|$ in (4).
To estimate the modulus of the second series, first observe that when $n\not\equiv 0\pmod{q}$, then
\begin{align*}
\left|1-z_{r}^{n}\right|^{2}&=\left(1-r^{n}\cos \frac{2\pi np}{q}\right)^{2}+r^{2n}\sin^{2}\frac{2\pi np}{q}\\
&=1-2r^{n}\cos\frac{2\pi np}{q}+r^{2n}\\
&=1+r^{2n}-2r^{n}\left(\cos^{2}\frac{\pi np}{q}-\sin^{2}\frac{\pi np}{q}\right)\\
&=\left(1-r^{n}\right)^{2}+4r^{n}\sin^{2}\frac{\pi np}{q}
\end{align*}
where we make use of the trigonometric identities $\cos^{2}\theta+\sin^{2}\theta=1$ and $\cos 2\theta=\cos^{2}\theta-\sin^{2}\theta$, for real $\theta$. It is not hard to verify that $\left|\sin\pi np/q\right|\geq\left|\sin\pi/q\right|$ for all $n\not\equiv 0 {\pmod q}$. Taking square roots of both sides, we conclude that
\begin{align*}
\left|1-z_{r}\right|\geq 2r^{n/2}\left|\sin \frac{\pi}{q}\right| \tag{5}
\end{align*}
By the triangle inequality,
\begin{align*}
\left|\left(1-r\right)\sum_{{n\geq 1}\atop {n\not\equiv 0\pmod{q}}}\dfrac{z_{r}^{n}}{1-z_{r}^{n}}\right|\leq \dfrac{\left(1-r\right)}{2}\sum_{n=0}^{\infty}\dfrac{r^{n}}{r^{n/2}\left|\sin \pi/q\right|}=\dfrac{1+r^{1/2}}{2\left|\sin\pi/q\right|}, \tag{6}
\end{align*}
which is uniformly bounded in $0<r<1$.
Combining results (4) and (6) together with the reverse triangle inequality, we conclude that
\begin{align*}
\left|f(z_{r})\right|\geq \dfrac{1}{q}\ln\left|\dfrac{1}{1-r^{q}}\right|-\dfrac{1}{\left|\sin\pi/q\right|}\rightarrow+\infty, \quad r\rightarrow 1^{-} \tag{7}
\end{align*}
