This question is from lecture 13 of the notes of Sieve Theory here:http://www.math.tau.ac.il/~rudnick/courses/sieves2015.html
I have 2 questions in the proof of lemma 2.2 on page 3:
Question 1 : I am not able to understand how should I prove that $\sum_{n\leq z} \frac{ \mu^2(d)} {\phi(d)}$ is bounded below by $ { log z} $.
Question 2 : In the line 6 of proof of lemma 2.2 how did the author removed the condition that gcd( h,q/l)=1?
I have thought a lot about these 2 problems but I am not able to prove these 2 questions.
Kindly help me.
For the first question, it suffices to use the properties of Euler product: \begin{aligned} \sum_{d\le z}{\mu^2(d)\over\varphi(d)} &=\sum_{d\le z}\mu^2(d)\prod_{p|d}{p^{-1}\over1-p^{-1}}=\sum_{d\le z}\mu^2(d)\prod_{p|d}\sum_{r\ge1}{1\over p^r} \\ &=\sum_{d\le z}\mu^2(d)\sum_{\substack{n\ge1\\p|n\Rightarrow p|d}}\frac1n=\sum_{n\ge1}\frac1n\sum_{\substack{d\le z\\p|n\iff p|d}}\mu^2(d) \\ &\ge\sum_{n\le z}\frac1n\sum_{\substack{d\le z\\d\text{ squarefree}\\p|d\iff p|n}}1\ge\sum_{n\le z}\frac1n. \end{aligned}
The last inequality follows from the fact that $d=\prod_{p|n}p$ is always an admissible option.
In the second question, the lecture note's author skipped some steps. Here is a complete version: \begin{aligned} \sum_{d\le z}{\mu^2(d)\over\varphi(d)} &=\sum_{\ell|q}\sum_{\substack{d\le z\\(d,q)=\ell}}{\mu^2(d)\over\varphi(d)}=\sum_{\ell|q}{\mu^2(\ell)\over\varphi(\ell)}\sum_{\substack{\ell h\le z\\(\ell h,q)=\ell \\(\ell,h)=1}}{\mu^2(h)\over\varphi(h)} \\ &=\sum_{\ell|q}{\mu^2(\ell)\over\varphi(\ell)}\sum_{\substack{\color{blue}{h\le z/\ell}\\(h,q)=1}}{\mu^2(h)\over\varphi(h)}\le\color{purple}{\sum_{\ell|q}{\mu^2(\ell)\over\varphi(\ell)}}\sum_{\substack{\color{red}{h\le z}\\(h,q)=1}}{\mu^2(h)\over\varphi(h)}, \end{aligned}
and writing the purple sum as a product over primes concludes the proof.