Vinogradov's Theorem/Major Arcs/Lemma 1

Lemma
Let $\phi$ be the Euler phi function, $\mu$ the Mobius function and $c_q$ the Ramanujan sum modulo $q$.

For $P,N \geq 1$, define:


 * $\displaystyle \mathfrak S_P(N) = \sum_{q \leq P}\frac{\mu(q)c_q(N)}{\phi(q)^3},\quad \mathfrak S(N) = \lim_{P \to \infty} \mathfrak S_P(N)$

Then $\mathfrak S(N) = \mathfrak S_P(N) + \mathcal O(P^{\epsilon -1})$ and $\mathfrak S$ has the Euler product:


 * $\displaystyle\mathfrak S(N) = \prod_{p \nmid N} \left( 1 + \frac 1{(p-1)^3} \right)\prod_{p \mid N} \left( 1 - \frac 1{(p-1)^2} \right)$

Proof
Firstly, we have

Trivially we have $|\mu(q)| \leq 1$. Therefore

Therefore by Convergence of Powers of Reciprocals $\mathfrak S(N) = \mathfrak S_P(N) + \mathcal O(P^{\epsilon -1})$ as claimed.

By Euler Product, and because $\mu(p^k) = 0$ for $k > 1$, we have:


 * $\displaystyle \mathfrak S(N) = \prod_p \left\{ 1 + \frac{\mu(q)c_q(N)}{\phi(q)^3} \right\}$

Now for a prime $p$ we have:


 * $\mu(p) = -1$


 * $\phi(p) = p - 1$

We also have Kluyver's Formula for Ramanujan's Sum


 * $\displaystyle c_p(n) = \sum_{d \backslash \gcd(p,n)} d \mu \left( \frac pd \right)$

If $p \mid N$ then $\gcd(p,N) = p$, and this gives


 * $c_p(N) = p\mu(1) + \mu(p) = p-1$

If $p \nmid N$ then $\gcd(p,N) = 1$, and $c_p(N) = -1$. Therefore


 * $\displaystyle \mathfrak S(N) = \prod_{p\mid N} \left\{ 1 - \frac{1}{(p-1)^2} \right\}\prod_{p\nmid N} \left\{ 1 + \frac{1}{(p-1)^3} \right\}$

This completes the proof.