<s>[INST] Verifying $|F(r)| \geq \frac{1}{1-r}\log(\frac{1}{1-r}) $ and $|F(re^{i \theta})| \geq c_{q/r}\frac{1}{1-r}\log({\log(\frac{1}{1-r})})$ I'm attempting to take a Tauberian route in verifying the proposition in $(1)$ below, which is from Complex Analysis, by Elias M Stein and Rami M. Shakarchi.
Let $F(z)$ be the following series:
$$F(z) = \sum_{n=1}^{\infty}d(n)z^{n} \, \, \text{for} \, |z| < 1$$
$\text{Remark}$
One can also observe the following relationship:
$$\sum_{n=1}^{\infty}d(n)z^{n} = \sum_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}$$
$(1)$
If $z=r$ with $0 < r < 1$, then as $r \rightarrow 1$, another case that be considered is $\theta=\frac{2\pi p}{q}$, where $p$ and $q$ are positive integers and then:
$(1.2)$
$$|F(r)| \geq \frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$
$(1.3)$
$$\lvert F(re^{i \theta})\rvert \geq c_{q/r}\frac{1}{1-r}\log\left(\frac{1}{1-r}\right)$$
$\text{Lemma}$
Formally attacking $(1)$, one can make the initial observations for the case seen in $(1.2)$
$$\left\lvert \sum_{n=1}^{\infty} \frac{z^{n}}{1-z^{n}}\right\rvert \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
$$\left\lvert\frac{z^{1}}{1-z^{1}} + \frac{z^{2}}{1-z^{2}} + \frac{z^{3}}{1-z^{3}} + \frac{z^{4}}{1-z^{4}} + \cdot \cdot \cdot + \frac{z^{n}}{1-z^{n}}\right\rvert \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
Recall the archetypal technique of Abel summability as formally developed in $(2)$:
$(2)$
$\text{Definition  (0.2)}$
A series $A(r)= \sum_{n=1}^{\infty}a_{n}r^{n}$ is said to Abel summable to $L$ if  $f(r)$ is convergent for all $\lvert r\rvert < L$ and if $f(r)$ converges to some limit $L$ as $r \rightarrow 1^{-}$:
$$A(r)= \sum_{n=1}^{\infty}a_{n}r^{n}$$
$\text{Remark}$: 
The developments of Abel summability, expressed within a prior definition, can be fully expressed as follows:
$$\lim_{r \rightarrow 1^{}} A(r)= \lim_{r \rightarrow 1^{}} \sum_{n=1}^{\infty}a_{n}r^{n} = L$$
Now recall our previous observation with some added developments:
$$\left|\frac{z^{1}}{1-z^{1}} + \frac{z^{2}}{1-z^{2}} + \frac{z^{3}}{1-z^{3}} + \frac{z^{4}}{1-z^{4}} + \cdot \cdot \cdot + \frac{z^{n}}{1-z^{n}} = d(1)z^{1}  + d(2)z^{2} + d(3)z^{3} + \cdot \cdot \cdot + d(4)z^{4} + d(z)z^{n}\right| \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
$$\left|d(1)z^{1}  + d(2)z^{2} + d(3)z^{3} + \cdot \cdot \cdot + d(4)z^{4} + d(z)z^{n}\right| \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
$$\lim_{r \rightarrow 1^{}}|d(1)z^{1}  + d(2)z^{2} + d(3)z^{3} + \cdot \cdot \cdot + d(4)z^{4} + d(z)z^{n}| \leq \frac{1}{1-r}\log\left( \frac{1}{1-r}\right)$$
Are the recent developments valid so far? I feel like this approach is too "archetypal." If the developments are wrong, may I have an answer on an alternate approach?
 [/INST] For the case of (1.2), note that
$$1-r^n = (1-r)(1+r+r^2+\cdots + r^{n-1}) \le (1-r)\cdot n.$$
Therefore
$$ \sum_{n=1}^{\infty}\frac{r^n}{1-r^n}\ge \sum_{n=1}^{\infty}\frac{r^n}{(1-r)n} = \frac{1}{1-r}\sum_{n=1}^{\infty}\frac{r^n}{n}.$$
But the sum on the right is precisely $\ln\left (\dfrac{1}{1-r}\right ).$ This gives your (1.2).
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