The sum $\sum_{n\leq x}\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2\pi \frac{k}{n}$ diverges as $x$, when $x$ tends to infitity

Question

I want to know if it is possible find an easy proof (this is without an use of an strong result) of

Question. Prove that the following sum diverges as $x\to\infty$ $$\sum_{n\leq x}\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2\pi \frac{k}{n},$$ optionally if it is possible, prove that diverges as $x$, or improve this result at infinity, without the use of a strong result seems Prime Number Theorem. Thanks in advance.

In the following appendix I show how we obtain previous sum, and the relationship with $M(x)+\sum_{n\leq x}\varphi(n)$, where $M(x)$ is the Mertens function and $\varphi(n)$ is Euler's totient function. Are only symbolic computations without a remarkable method. Suggestions are welcome.

Appendix: We have the following fact

Fact 1. Let $n\geq 1$ an integer, $\varphi(n)$ is Euler's function $\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}1$, and $\mu(n)$ is the Möbius function, defined by $\mu(1)=1$, $\mu(n)=(-1)^k$ if $n$ is a product of $k$ distinct primes, and is equals to $0$ otherwise. Then $$\varphi(n)+\mu(n)=2\left(\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2\pi \frac{k}{n}\right).$$

Proof. Since $\mu(n)$ can be written as $\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}e^{2\pi i\frac{k}{n}}$ (see [1] for example), and $\mu(n)\in\{-1,0,1\}$, this is a real function, thus taking the real part previous equations gives $$\mu(n)=\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos 2\pi\frac{k}{n}.$$ Now too it is well known the trigonometric identity $\cos2\theta=2\cos^2\theta-1$. Fixed an integers $n\geq 1$, putting for each term $\theta_k=\pi\frac{k}{n}$, we obtain the desired result when we add all terms over $1\leq k\leq n$ such that $gcd(k,n)=1$, and this completes the proof.

Example. The quantity $\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2 \pi\frac{k}{n}$ isn't always an integer, for example when $n=6$, $\varphi(n)=2$, $\cos\frac{5\pi}{6}=-\cos\frac{\pi}{6}=-\frac{\sqrt{3}}{2}$, and we obtain $\cos^2\frac{\pi}{6}+\cos^2\frac{5\pi}{6}=\frac{3}{2}$.

The mean $\sum_{n\leq x}\mu(n)$ for Möbius function is called Mertens function and denoted by $M(x)$.

Fact 2. We can state a equation between means if add terms $n\leq x$ in equation of previous $$M(x)+\sum_{n\leq x}\varphi(n)=2\sum_{n\leq x}\left(\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2\pi \frac{k}{n}\right).$$

Now, we know the following theorems (see [2]):

Theorem 1. For $x>1$ $$\sum_{n\leq x}\varphi(n)=\frac{3}{\pi^2}x^2+O(x\log x).$$

Theorem 2. The Prime Number Theorem implies (in fact, is equivalent to) $\lim_{x\to\infty}\frac{M(x)}{x}=0$.

Thus using Fact 2 with previous theorems we compute easily

$$\infty=0+\lim_{x\to\infty}\frac{1}{x}\sum_{n\leq x}\varphi(n)=\lim_{x\to\infty}2\sum_{n\leq x}\left(\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2\pi \frac{k}{n}\right),$$ and we can claim that the series $\sum_{n=1}^{\infty}\left(\sum_{\substack{1\leq k\leq n \\ gdc(k,n)=1}}\cos^2\pi \frac{k}{n}\right)$ diverges as $x$, when $x\to\infty$. Can we find a proof without the use of these set of theorems, specially the use concerning to Prime Number Theorem?

I too tried combine this 'method' with the trigonometric identity $3+4\cos\theta+\cos 2\theta=2(1+\cos\theta)^2$, but I don't find a way to obtain something useful.

References:

[1] Ram Murty, Problems in Analytic Number Theory, Springer 2008. Exercise 1.1.13, pages 206-207.

[2] Apostol, Introduction to Analytic Number Theory , Springer 1976. Theorem 3.7 in page 62 and Theorem 4.14 in pages 92-94.

Answer 1

Here's a stupid proof that it diverges: we have that \[\sum_{\substack{1 \leq k \leq n \\ \gcd(k,n) = 1}} \cos^2 \frac{\pi k}{n} \geq \cos^2 \frac{\pi}{n}, \] simply by discarding all the terms other than $k = 1$ and using the fact that $\cos^2 x$ is nonnegative. So the sum you are interested in is bigger than \[\sum_{n \leq x} \cos^2 \frac{\pi}{n}.\] For $n \geq 4$, say, \[0 \leq \frac{\pi}{n} \leq \frac{\pi}{4},\] and as $\cos^2 x$ is decreasing on the interval $[0,\pi/4]$, it follows that the sum is bigger than \[\sum_{4 \leq n \leq x} \cos^2 \frac{\pi}{4} = \sum_{4 \leq x} \frac{1}{2} = \frac{1}{2} \left(\lfloor x \rfloor - 3\right).\] Clearly this diverges as $x \to \infty$.

Of course, the other method you use actually gives an asymptotic expression for this sum, which is much nicer.