# Lower bound for $F(z) = \sum_{n=1}^\infty d(n)z^n$ near radius of convergence.

In Stein and Sharkarchi Problem 2.7.2 one is asked to find a lower bound $$|F(z)| \geq c\frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$ for the function $$F(z) = \sum_{n=1}^\infty d(n)z^n$$ near the radius of convergence $R$. Here $d(n)$ is the number of divisors of $n$. I have already established $R = 1$ and the identity $$\sum_{n=1}^\infty d(n)z^n = \sum_{n=1}^\infty \frac{z^n}{1-z^n}.$$

Furthermore, when $z$ is a real number $r$ then clearly $$F(r) = \sum_{n=1}^\infty \frac{r^n}{1-r^n} \geq \int_1^\infty \frac{r^n}{1-r^n} dn = -\frac{1}{\log(r)}\log\left(\frac{1}{1-r}\right)$$ by substituting $u = 1-r^n$ and integrating. Since $r \to 1$ I have $\log(r) = (r-1) + O((r-1)^2) \approx (r-1)$ and thus $$F(r) \geq c\frac{1}{1-r}\log\left(\frac{1}{1-r}\right),$$ for some constant $c$ coming from the approximation of $\log(r) \approx (r-1)$.

In the problem I have to establish the same lower bound when $z = re^{i\theta}$ for some $\theta = 2\pi p/q$ with $p$ and $q$ integers. Can I somehow use a similar approach for this? or do I need to do something completely different?

Edit #1: added difinition of $d(n)$.

Edit #2: Okay, I think I have an answer now. Maybe someone can verify or deny it.

Let $N_q := \{n\in \mathbb{N}\mid q\ \text{does not divide}\ n \}$ and consider the follwing splitting of the sum $$|F(z)| = \left|\sum_{n=1}^\infty \frac{z^n}{1-z^n}\right| = \left|\underbrace{\sum_{n=1}^\infty\frac{z^{qn}}{1-z^{qn}}}_{=: A} + \underbrace{\sum_{n\in N_q}\frac{z^n}{1-z^n}}_{=: B}\right|.$$

Now part $A$ is strictly real so with $z = r$ we have as before $$A \geq c_A\frac{1}{1-r^q}\log\left(\frac{1}{1-r^q}\right) \geq c_A\frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$ since $r < 1$.

For part $B$ we have th following observation. Since $e^{ik\theta} = e^{ik\theta}e^{iq\theta} = e^{i(k+q)\theta}$ clearly we only have $q-1$ directions to worry about and let $k'$ be defined such that $e^{ik'\theta}$ is the closest to 1 when following the unitcicle perimiter. Let $c_B = \min\{1,\inf\{|1-re^{ik'\theta}|\mid 0< r <1 \}\}$ then we have $$|B| \leq \sum_{n\in N_q}\frac{|z|^n}{|1-z^n|} \leq \sum_{n\in N_q}\frac{|z|^n}{c_B} \leq \frac{c_B^{-1}}{1-|z|} = \frac{c_B^{-1}}{1-r}$$.

Now, since $A \geq 0$ we have $|A+B|\geq A-|B|$ and thus $$F(z) \geq A - |B| \geq c_A\frac{1}{1-r}\log\left(\frac{1}{1-r}\right) - |B| \geq c_A\frac{1}{1-r}\log\left(\frac{1}{1-r}\right) - \frac{c_B^{-1}}{1-r},$$ but this we can rewrite at bit $$c_A\frac{1}{1-r}\log\left(\frac{1}{1-r}\right) - \frac{c_B^{-1}}{1-r} = c_A\frac{1}{1-r}\left(\log\left(\frac{1}{1-r}\right) - c_B^{-1}\right)$$ and since $\log\left(\frac{1}{1-r}\right) \gg c_B^{-1}$ as $r \to 1$ we simply have $$F(z) \geq c_{p/q}\frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$ for some constant $c_{p/q}$.

Edit #3: I realize that there was something wrong. I should probably assume $p$ and $q$ relatively prime.

• Maybe it would be nice to explain what $d(n)$ is. It would also increase the probability that you get answers ;-).
– Karl
Sep 30, 2014 at 12:14
• Ah right, $d(n)$ is the number of divisors of $n$, I actually thought that was a standard notation, and given all the other information about the series, I didn't find that excessively important. I'll edit it in, thanks.
– zo0x
Sep 30, 2014 at 12:35
• Why exactly is it clear that $F(r) = \sum_{n=1}^\infty \frac{r^n}{1-r^n} \geq \int_1^\infty \frac{r^n}{1-r^n} dn?$
– user225477
Mar 6, 2017 at 0:08
• @Zermelo's_Choice as r<1 the integrant is decreasing as n increases. If you think of the sum as an integral over the simple function where you round n down. This staircase graph will major the graph of the continuous integrant.
– zo0x
Mar 7, 2017 at 5:58
• $F(r) = \sum_{n=1}^\infty \frac{r^n}{1-r^n} \geq \int_1^\infty \frac{r^n}{1-r^n} dn?$ looks like he's applying the integral test, I tried solving the same problem here:math.stackexchange.com/questions/2355465/… Jul 17, 2017 at 16:14