TITLE: Proving $\sum_{k=1}^n\vert\cos(k)\vert\ge \frac{n}{4}$ for all $n>0$
QUESTION [6 upvotes]: Show that for all $n\in\mathbb{N}^* \quad \sum_{k=1}^n\vert\cos(k)\vert\ge \frac{n}{4}.$

I can do it using the fact that $\vert\cos(x)\vert\ge \cos(x)^2$
\begin{equation}
\sum_{k=1}^n\vert\cos(k)\vert\ge \sum_{k=1}^n\cos(k)^2=\frac{n}{2}+\frac1{2}\Re\big(e^{2i}\frac{1-e^{2in}}{1-e^{2i}}\big)\\
=\frac{n}{2}+\frac{1}{2}\Re\big(e^{i(n+1)}\frac{\sin(n)}{\sin{1}}\big)=\frac{n}{2}+\frac{\cos(n+1)\sin(n)}{2\sin(1)}\ge\frac{n}{2}-\frac{1}{2\sin(1)}
\end{equation}
I can finish by using "calculator" or by hand but I would rather like a "direct" proof not using the value of $\sin(1).$
Other methods can be very interesting too.

REPLY [2 votes]: Since
$$ \left|\cos x\right|=\frac{2}{\pi}+\frac{4}{\pi}\sum_{m\geq 1}\frac{\cos(2mx)(-1)^{m+1}}{4m^2-1} $$
and
$$ \left|\sum_{n=1}^{N}\cos(2mn)\right|\leq \frac{1}{\left|\sin m\right|}\leq \frac{\pi/2}{d(m,\pi\mathbb{Z})} $$
we have
$$ \left|-\frac{2n}{\pi}+\sum_{k=1}^{n}\left|\cos k\right|\right|\leq \frac{1}{2}\sum_{m\geq 1}\frac{1}{(4m^2-1)\,d(m,\pi\mathbb{Z})}. \tag{A}$$
Once we prove that the series in the RHS of $(A)$ is convergent we have $\sum_{k=1}^{n}\left|\cos k\right|=\frac{2n}{\pi}+O(1)$, which is way stronger than the original inequality. Now $m^2\,d(m,\pi\mathbb{Z})$ may be as small as $\Theta(m)$, but by Lagrange's theorem that happens iff $m$ is the numerator of a convergent of $\pi$, i.e. an element of the sequence $\{\eta_n\}_{n\geq 1}=\{3,22,333,355,103993,\ldots\}$. Since $\pi$ is irrational this sequence has at least an exponential growth and the series $\sum_{n\geq 1}\frac{1}{\eta_n}$ is convergent. As a consequence, the RHS of $(A)$ is convergent too. An approximated value for it is $0.6$. Here it is a graph of the LHS of $(A)$ for $n\in[1,1000]$:
$\hspace{1cm}$