Ramanujan's sum bounded

Question

Is there a simple proof why Ramanujan's sum $\vert c_n(a) \vert \leqslant a$, when $a \in \mathbb N$. I found this statement, but don't know how to prove it.

Answer 1

The reason for the limitation can be seen with the formula

$$c_n(a):=\sum\limits_{k=1,(k,n)=1}^n \exp(i2\pi \frac{ak}{n})=\varphi(n)\frac{\mu(\frac{n}{gcd(n,a)})}{\varphi(\frac{n}{gcd(n,a)})} \enspace.$$

Be $\enspace\displaystyle n=\prod\limits_{k=1}^r p_k^{a_k}\enspace$ and $\enspace\displaystyle gcd(n,a)=\prod\limits_{k=1}^s p_k^{b_k}\enspace$ ordered by common prime factors with $\enspace s\leq r\enspace$, $\enspace 1\leq b_k\leq a_k\enspace$ and $\enspace\displaystyle c_k:=(1|_{a_k=b_k} \vee 0|_{a_k\neq b_k})$ .

It follows $$c_n(a)=\mu(\frac{n}{gcd(n,a)}) gcd(n,a)\prod\limits_{k=1}^r(1-\frac{1}{p_k})^{1-c_k}$$
and therefore $\enspace|c_n(a)|\leq gcd(n,a)\enspace$.