- \(^\ast\) Let
\[ F (z) = \sum_ {n = 1} ^ {\infty} d (n) z ^ {n} \quad \mathrm{for} | z | < 1 \]
where \(d ( n )\) denotes the number of divisors of \(n .\) Observe that the radius of convergence of this series is 1. Verify the identity
\[ \sum_ {n = 1} ^ {\infty} d (n) z ^ {n} = \sum_ {n = 1} ^ {\infty} \frac {z ^ {n}}{1 - z ^ {n}}. \]
Using this identity, show that if \(z = r\) with \(0 < r < 1\) , then
\[ | F (r) | \geq c \frac {1}{1 - r} \log (1 / (1 - r)) \]
as \(r \to 1\) . Similarly, if \(\theta = 2 \pi p / q\) where \(p\) and \(q\) are positive integers and \(z = r e ^ { i \theta }\) then
\[ | F (r e ^ {i \theta}) | \geq c _ {p / q} \frac {1}{1 - r} \log (1 / (1 - r)) \]
Solution
1. \(\sum_{n\ge1} d(n) z^n = \sum_{n\ge1} \frac{z^n}{1 - z^n}\) for \(\abs z<1\).
By the geometric series, \(\frac{z^n}{1 - z^n} = \sum_{k\ge1} z^{kn}\). The double series converges absolutely, and the coefficient of \(z^m\) in \(\sum_n\sum_k z^{kn}\) is the number of pairs \((n,k)\) with \(kn = m\), which is \(d(m)\).
2. \(F(r) \ge c \frac{1}{1-r}\log\frac{1}{1-r}\) as \(r \to 1^-\).
All terms of \(F(r) = \sum_{n} \frac{r^n}{1-r^n}\) are positive. Using \(1 - r^n \le n(1-r)\) and keeping the terms with \(n \le N\), where \(N\) is the integer part of \(\frac{1}{1-r}\), gives \(F(r)\ge\frac{r^N}{1-r}\sum_{n\le N}\frac1n\). Since \(r^N\) is bounded below by a positive constant as \(r\to1^-\) and \(\sum_{n\le N}\frac1n\ge\log N\), this is at least \(c \frac{1}{1-r}\log\frac{1}{1-r}\).
3. For \(\theta = 2\pi p/q\), \(\abs{F(re^{i\theta})} \ge c_{p/q}\frac{1}{1-r}\log\frac{1}{1-r}\) as \(r\to1^-\).
Assume \(p/q\) is in lowest terms and put \(z=re^{i\theta}\). For \(q\mid n\), \(z^n = r^n\), and the terms with \(n=qj\) contribute \(\sum_j\frac{r^{qj}}{1-r^{qj}}=F(r^q)\), which by step 2 is at least \(c\frac{1}{1-r^q}\log\frac{1}{1-r^q}\ge \frac cq\frac{1}{1-r}\log\frac{1}{q(1-r)}\) since \(1-r^q\le q(1-r)\). For \(q\nmid n\), \(e^{in\theta}\) is a \(q\)th root of unity different from \(1\), so \(\abs{1-z^n}\ge\delta_q>0\) for \(r\) near \(1\), and these terms contribute at most \(\sum_n r^n/\delta_q=O\bigl(\frac1{1-r}\bigr)\). The logarithmic factor dominates, which gives the bound with a smaller constant \(c_{p/q}\).