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.)

  • $\begingroup$ Hint: What happens to your function near roots of unity? $\endgroup$ – BigMathTimes Jul 2 '15 at 16:03
  • $\begingroup$ If $|z|=1$ then we can write $z= e^{i x}$ for some real $x$. Show that $\frac{z^n}{1-z^n} = \frac{1}{2}\frac{\cos(nx) - \cos(2nx)}{1 - \cos(nx)} + \frac{i}{2}\frac{\sin(nx) - \sin(2nx)}{1 - \cos(nx)} = a_n + i b_n$ and note that the sum converge if and only if both $\sum a_n$ and $\sum b_n$ converge. A useful fact: if $\sum a_n$ converge then we must have $\lim_{n\to\infty} a_n = 0$. Is this the case for any $x$? $\endgroup$ – Winther Jul 2 '15 at 16:16
  • 1
    $\begingroup$ @Winther Yes the series diverges at evey point on the boundary. That's also true of $f(z) = \sum z^n.$ But $f(z) = 1/(1-z)$ in the disc, so $f$ extends to be holomorphic in $\mathbb C \setminus \{1\}.$ Thus a badly behaved series at every point on the boundary does not imply it's a natural boundary for that series. $\endgroup$ – zhw. Jul 3 '15 at 3:47
  • $\begingroup$ @zhw. Yes, I misread the question, thought it was about convergence. Thanks for pointing it out. $\endgroup$ – Winther Jul 3 '15 at 9:06
  • $\begingroup$ @BryanBrown: Would you please elaborate? I tried playing around with that suggestion, but didn't get anywhere (except trivially for $z=1$). $\endgroup$ – Matt Rosenzweig Jul 3 '15 at 14:22

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*}


\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*}


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