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Problem: (Davenport 1932) Let \[f(n) = -\sum_{d|n} \frac{\mu(d)\log d}{d}.\]

(a) By recalling Exercise 2.1.16(a), or otherwise, show that \(f(n) \ge 0\) for all \(n\).

(b) Show that \(f(n) \ll \log\log n\) for \(n \ge 3\).

(c) Show that \(f(n) \sim \frac{1}{4}\log\log n\) if \(n = \prod_{y \lt p \le y^2}p\).

(d) Show that \(f(n) \le\left( \frac{1}{4} + o(1)\right)\log\log n\) as \(n\to \infty\).

Solution: (a) By Exercise 2.1.16(a), \[f(n) = -\sum_{d|n} \frac{\mu(d)\log d}{d} = \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p - 1} \ge 0.\]

(b) Note that \begin{align*} f(n) \le \sum_{p \mid n} \frac{\log p}{p - 1}. \end{align*} Since \({\log x}/{(x - 1)}\) is decreasing function for all \(x\gt 0\), it is enough to show only when \(n = \prod_{p \le y}p\). Now, \begin{align*} \sum_{p \mid n} \frac{\log p}{p - 1} &= \sum_{p \le y} \frac{\log p}{p-1}\\ &\ll \log y = \log \log n. \end{align*}

(c) \begin{align*} f(n) &= \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p - 1}\\ &= \prod_{y \lt p \le y^2} \left(1-\frac{1}{p} \right) \sum_{y \lt p \le y^2} \frac{\log p}{p-1}\\ &= \left(\frac{1}{2} + o(1)\right) \left(\log y + O(1) \right)\\ &= \left(\frac{1}{4} + o(1)\right)\log\log n \end{align*}

(d) Suppose \(n = \prod_{y^{a} \lt p \le y^2}p\), then \begin{align*} f(n) &= \frac{\varphi(n)}{n} \sum_{p \mid n} \frac{\log p}{p - 1}\\ &= \prod_{y^{a} \lt p \le y^2} \left(1-\frac{1}{p} \right) \sum_{y^{a} \lt p \le y^2} \frac{\log p}{p-1}\\ &= \left(\frac{\log y^{a}}{2\log y} + o(1)\right) \left(\log y^2 - \log y^{a} + O(1) \right)\\ &= \left(\frac{a}{2} + o(1)\right) \left((2-a)\log y+ O(1) \right)\\ &= \left(\frac{a(2-a)}{4} + o(1)\right)\log\log n\le\left( \frac{1}{4} + o(1)\right)\log\log n. \end{align*}

Problem: Let \(d_k (n)\) be as in Exercise 2.1.18. Show that if \(k\) and \(\kappa\) are fixed, then \[\sum_{n\le x} d_k (n) ^{\kappa} \ll x (\log x) ^{k ^{\kappa}-1}.\] for \(x \ge 2\).

Solution: Since \[\sum_{p \le x}d_{k}(p)^{\kappa}\log p \ll x, \qquad \sum_{\substack{p^{m}\\m\ge 2}}\frac{d_{k}(p^{m})^{\kappa}m\log p}{p^{m}} \ll 1,\] by Corollary 2.15, \begin{align*} \sum_{n\le x} d_k (n) ^{\kappa} &\ll \frac{x}{\log x} \prod_{p \le x} \left(1 + \frac{d_k (p) ^{\kappa}}{p}+\frac{d_k (p^2) ^{\kappa}}{p^2}+\cdots \right)\\ &\ll x (\log x) ^{k ^{\kappa}-1}. \end{align*}

Problem: Let \(f(n) = \prod_{p|n}(1+p^{-1/2})\).

(a) Show that there is a constant \(a\) such that if \(n\ge 3\), then \[f(n) \lt \exp (a(\log n)^{1/2}(\log \log n)^{-1}). \]

(b) Show that \(\sum_{n\le x}f(n) = cx + O(x^{1/2})\) where \(c = \prod_{p}(1+p^{-3/2})\).

Solution: (a) It is enough to show only when \(n = \prod_{p\le y} p\). Note that \(\log \log n = \log y + O(1)\) and \[\sum_{p\le y}p^{-1/2} = \int_{2-}^{y} u^{-1/2} d\pi(u) \ll y^{1/2}(\log y)^{-1}.\] Now, \begin{align*} \log f(n) &= \sum_{p\le y}\log(1+p^{-1/2})\\ &=\sum_{p\le y}p^{-1/2} + O \left(\sum_{p\le y}p^{-1}\right)\\ &\ll y^{1/2}(\log y)^{-1} = (\log n)^{1/2} (\log \log n)^{-1}. \end{align*}

(b) Note that \[f(n) = \sum_{d|n} \mu(d)^2 d^{-1/2}.\] Hence, \begin{align*} \sum_{n\le x}f(n) &= \sum_{n\le x}\sum_{d|n} \mu(d)^2 d^{-1/2}\\ &=x\sum_{d\le x}\mu(d)^2 d^{-3/2} + O\left(\sum_{d\le x} \mu(d)^2 d^{-1/2}\right)\\ &= x\sum_{d=1}^{\infty}\mu(d)^2 d^{-3/2} + \sum_{d \gt x}\mu(d)^2 d^{-3/2} + O(x^{1/2})\\ &= cx + O(x^{1/2}). \end{align*}

Problem: Show that \(d(n) \le \sqrt{3n}\) with equality if and only if \(n=12\).

Solution: Follow the steps in proving (2.20). With \(n = \prod_p p^{a_p}\), \[\frac{d(n)}{n^{1/2}} = \prod_{p}\frac{a_p + 1}{p^{a_p / 2}},\] which be maximized by \(a_p = [(p^{1/2}-1)^{-1}].\) Since \(a_2 = 2\), \(a_3 = 1\), and \(a_p = 0\) for \(p \ge 5\), \(d(n) \le \sqrt{3n}\), where equality holds when \(n = \prod_p p^{a_p} = 12\).

Problem: Let \(\sigma(n) = \sum_{d|n}d\).

(a) Show that \(\sigma(n) \varphi(n) \le n^2\) for all \(n \ge 1\).

(b) Deduce that \(n+1 \le \sigma(n) \le e^{C_0}n\big(\log \log n + O(1)\big)\) for all \(n\ge 3\).

Solution: (a) Note that \[\sigma(p^a) \varphi(p^a) = p^{2a} - p^{a-1} \le p^{2a} .\] Hence, for \(n = p_{1}^{\alpha_1} p_{2}^{\alpha_2} \cdots p_{k}^{\alpha_k}\), \[\sigma(n) \varphi(n) = \sigma({p_{1}^{\alpha_1}})\varphi(p_{1}^{\alpha_1}) \cdots \sigma({p_{k}^{\alpha_k}})\varphi(p_{k}^{\alpha_k}) \le p_{1}^{2\alpha_1} \cdots p_{k}^{2\alpha_k} = n^2.\]

(b) By Theorem 2.9, we're done.