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I am trying to show that if $$F(z) = \sum_{n=1}^{\infty}\frac{z^n}{1-z^n}$$

then

$$|F(r)| \geq \frac{c}{1-r}log(\frac{1}{1-r})$$ as r -> 1 where $r \in (0,1)$

I honestly have no idea where to start.

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One might start from noticing that for any $z\in(0,1)$ we have $$ F(z)=\sum_{n\geq 1} d(n) z^n $$ by expanding $\frac{z^n}{1-z^n}$ as a geometric series and rearranging the series. The behaviour of the divisor function $d(n)$ is quite erratic, but there is a simple trick for mitigating it: $$ \frac{1}{1-z}F(z) = \sum_{n\geq 1}\left(\sum_{m=1}^{n}d(m)\right)z^n $$ and by Dirichlet's hyperbola method: $$ \sum_{m=1}^{n} d(m) = n H_n + (\gamma-1) n + O(\sqrt{n}) $$ where $$ \sum_{n\geq 1}\left(n H_n +(\gamma-1) n \right)z^n = \frac{\gamma z-z\log(1-z)}{(1-z)^2}$$ $$ \sum_{n\geq 1}\frac{n\sqrt{\pi}\binom{2n}{n}}{4^n}z^n = \frac{z\sqrt{\pi}}{2(1-z)^{3/2}},\qquad \frac{n\sqrt{\pi}\binom{2n}{n}}{4^n}\sim\sqrt{n} $$ ensure that in a left neighbourhood of $z=1$ we have $$ F(z) \sim \frac{-\log(1-z)+\gamma}{(1-z)}. $$

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