<s>[INST] Complex power series in Stein's book Let $$ F(z)=\sum\limits_{n=1}^{\infty}d(n)z^{n}$$for $ \left|z\right|<1 $,where $d(n)$ denotes the number of divisors of $ n $.Obeserve that the radius of convergence of this series is 1.Verify the identity $$\sum\limits_{n=1}^{\infty}d(n)z^{n}=\sum\limits_{n=1}^{\infty}\frac{z^{n}}{1-z^{n}} $$Using the identity,show that if $ z=r $ with $ 0<r<1 $,then $$\left|F(r)\right|\geqslant c\frac{1}{1-r}log(\frac{1}{1-r})$$as $ r\longrightarrow1 $.Similarly,if $ \theta=2\pi p/q $ where $ p $ and $ q $ are positive integers and $ z=re^{i\theta} $,then$$ \left|F(re^{i\theta})\right|\geqslant c_{p/q}\frac{1}{1-r}log(\frac{1}{1-r}) $$as $ r\longrightarrow1 $.Conclude that $ F $ cannot be continued analytically past the unit disc. I solve the problem when $ z=r $ is real,my method is here: $$\left|F(r)\right|=\left|\sum\limits_{n=1}^{\infty}\frac{r^{n}}{1-r^{n}}\right|=\left|\frac{1}{1-r}\right|\left|\sum\limits_{n=1}^{\infty}\frac{r^{n}}{1+r+\cdots+r^{n-1}}\right|\geqslant\frac{1}{1-r}\sum\limits_{n=1}^{\infty}\frac{r^{n}}{n}\geqslant c\frac{1}{1-r}log(\frac{1}{1-r})$$But when $ z=re^{i\theta} $ i found my method didn't work.So how can I solve the problems?
 [/INST] Let be $
z=z\left( r \right) = re^{2\pi i\frac{p}
{q}} 
$
We will prove that
$$
\mathop {\lim }\limits_{r \to 1^ -  } \left| {\left( {1 - r} \right)F\left( {z\left( r \right)} \right)} \right| =  + \infty 
$$
As first we write
$$
F(z) = \sum\limits_{n = 1}^{ + \infty } {\frac{{z^n }}
{{1 - z^n }} = \sum\limits_A {\frac{{z^n }}
{{1 - z^n }} + } } \sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} 
$$
where 
$$
A = \left\{ {n \in \mathbb{N}:n \equiv 0\,\,(\bmod \,\,q)} \right\}
$$
and
$$
B = \left\{ {n \in \mathbb{N}:n\not  \equiv 0\,\,(\bmod \,\,q)} \right\}
$$
We will prove that
$$
\mathop {\lim }\limits_{r \to 1^ -  } \left| {\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}} } \right| =  + \infty 
$$
and that 
$$
\left| {\left( {1 - r} \right)\sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} } \right| \leqslant M
$$
where $M$ is a suitable constant.
We note that if $
n \equiv 0\,(\bmod \,\,q)
$ then we can put $n=kq$ with $k$ integer and greater than $0$. Therefore
$$
z=\left( {z\left( r \right)} \right)^n  = \left( {re^{2\pi i\frac{p}
{q}} } \right)^{kq}  = r^{kq} 
$$
thus
$$
\begin{gathered}
  \left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}}  = \left( {1 - r} \right)\sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} }}
{{1 - r^{kq} }}}  =  \hfill \\
   \hfill \\
   = \frac{{\left( {1 - r} \right)}}
{{1 - r^q }}\sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} \left( {1 - r^q } \right)}}
{{1 - r^{kq} }}}  =  \hfill \\
   \hfill \\
   = \frac{1}
{{1 + r +  \cdots r^{q - 1} }} \cdot \sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} \left( {1 - r^q } \right)}}
{{\left( {1 - r^q } \right)\left( {1 + r^q  +  \cdots r^{q\left( {k - 1} \right)} } \right)}}}  =  \hfill \\
   \hfill \\
   = \frac{1}
{{1 + r +  \cdots r^{q - 1} }} \cdot \sum\limits_{k = 1}^{ + \infty } {\frac{{r^{kq} }}
{{\left( {1 + r^q  +  \cdots r^{q\left( {k - 1} \right)} } \right)}}}  \hfill \\ 
\end{gathered} 
$$
But $1+r+...r^{q-1} \leq q$ as well as $1+r^q+...r^{q(k-1)}\leq k$. Thus
$$
\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}}  \geqslant \frac{1}
{q}\sum\limits_{k = 1}^{ + \infty } {\frac{{\left( {r^q } \right)^k }}
{k}}  = \frac{1}
{q}\log \left( {\frac{1}
{{1 - r^q }}} \right)
$$
so that
$$
\mathop {\lim }\limits_{r \to 1^ -  } \left[ {\left( {1 - r} \right)\sum\limits_A {\frac{{z^n }}
{{1 - z^n }}} } \right] \geqslant \mathop {\lim }\limits_{r \to 1^ -  } \left[ {\frac{1}
{q}\log \left( {\frac{1}
{{1 - r^q }}} \right)} \right] =  + \infty 
$$
Now we consider 
$$
\left| {\left( {1 - r} \right)\sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} } \right|
$$
We have that 
$$
\begin{gathered}
  \left| {1 - z^n } \right|^2  = \left| {1 - z^n } \right|\left| {1 - \bar z^n } \right| =  \hfill \\
   \hfill \\
   = 1 - 2r^n \cos \frac{{2\pi pn}}
{q} + r^{2n}  =  \hfill \\
   \hfill \\
   = \left( {1 - r^n } \right)^2  + 2r^n \left( {1 - \cos \frac{{2\pi pn}}
{q}} \right) \geqslant  \hfill \\
   \hfill \\
   \geqslant 2r^n \left( {2\sin ^2 \frac{{\pi pn}}
{q}} \right) = 4r^n \sin ^2 \frac{{\pi pn}}
{q} \hfill \\ 
\end{gathered} 
$$
Now, let be
$$
np = aq + b\,\,\,\,\,0 < b < q
$$
It is
$$
\frac{{\pi pn}}
{q} = a\pi  + \frac{{\pi b}}
{q}
$$
so
$$
\sin ^2 \frac{{\pi pn}}
{q} = \sin ^2 \frac{{\pi b}}
{q} \geqslant \sin ^2 \frac{\pi }
{q}
$$
Hence,
$$
\left| {1 - z^n } \right|^2  \geqslant 4 r^n \sin ^2 \frac{\pi }
{q}
$$
and so
$$
\left| {1 - z^n } \right| \geqslant 2 r^{\frac{n}
{2}} \sin \frac{\pi }
{q}
$$
It follows that
$$
\begin{gathered}
  \left| {\left( {1 - r} \right)\sum\limits_B {\frac{{z^n }}
{{1 - z^n }}} } \right| \leqslant \left( {1 - r} \right)\sum\limits_{n = 0}^{ + \infty } {\frac{{\left| z \right|^n }}
{{\left| {1 - z^n } \right|}}}  \leqslant  \hfill \\
   \hfill \\
   \leqslant \frac{{\left( {1 - r} \right)}}
{{2\sin \frac{\pi }
{q}}}\sum\limits_{n = 0}^{ + \infty } {\frac{{r^n }}
{{r^{\frac{n}
{2}} }}}  = \frac{{\left( {1 - r} \right)}}
{{2\sin \frac{\pi }
{q}}}\sum\limits_{n = 0}^{ + \infty } {r^{\frac{n}
{2}} }  =  \hfill \\
   \hfill \\
   = \frac{{\left( {1 - r} \right)}}
{{2\sin \frac{\pi }
{q}}}\frac{1}
{{1 - r^{\frac{1}
{2}} }} = \frac{{\left( {1 + r^{\frac{1}
{2}} } \right)}}
{{2\sin \frac{\pi }
{q}}} \leqslant \frac{2}
{{2\sin \frac{\pi }
{q}}} = M \hfill \\ 
\end{gathered} 
$$
From this it follows that
$$
\mathop {\lim }\limits_{r \to 1^ -  } \left| {\left( {1 - r} \right)F(z\left( r \right))} \right| =  + \infty 
$$
which is even more of what you need.
</s>