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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.

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  • $\begingroup$ Maybe it would be nice to explain what $d(n)$ is. It would also increase the probability that you get answers ;-). $\endgroup$
    – Karl
    Sep 30, 2014 at 12:14
  • $\begingroup$ 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. $\endgroup$
    – zo0x
    Sep 30, 2014 at 12:35
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    $\begingroup$ Why exactly is it clear that $F(r) = \sum_{n=1}^\infty \frac{r^n}{1-r^n} \geq \int_1^\infty \frac{r^n}{1-r^n} dn?$ $\endgroup$
    – user225477
    Mar 6, 2017 at 0:08
  • $\begingroup$ @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. $\endgroup$
    – zo0x
    Mar 7, 2017 at 5:58
  • $\begingroup$ $F(r) = \sum_{n=1}^\infty \frac{r^n}{1-r^n} \geq \int_1^\infty \frac{r^n}{1-r^n} dn?$ looks like he's applying the integral test, I tried solving the same problem here:math.stackexchange.com/questions/2355465/… $\endgroup$
    – Zophikel
    Jul 17, 2017 at 16:14

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