An Inequality for a Trigonometric Sum
Using the equirepartion of the sequence $(n \mod 2\pi)$ one can show that
$$\lim_{n\to\infty} \frac1n \sum_{k=0}^n|\cos k|=\frac{2}{\pi}$$
Numerical evidence shows that, for every $n$,
$$ \sum_{k=0}^n|\cos k|>\frac{2}{\pi}n.$$
Can someone help proving this?
Indeed by numerical evaluation (up to $n = 500,000,000$) we have $$ s(n) = \sum_{k=0}^n|\cos k| - \frac{2}{\pi}n > 0.4$$
There are two obvious problems:
The sum $\sum_{k=0}^n|\cos k|$ approaches $1 + \int_{x=0}^n|\cos x| dx$, where alternatingly terms $|\cos k|$ in the sum are either overestimating or underestimating $\int_{x=k-1}^k|\cos x| dx$. Note for the "continuous" case, $1 + \int_{x=0}^y|\cos x| dx - \frac{2}{\pi}y > 0.78; \, \forall y$, with periodicity in $\pi$ of these characteristics.
$n$ will never be a multiple of $\pi$ so we will have no true periodicity.
We will see that
ad 1.: The effects of overestimating and understimating cancel out over very large intervals.
ad 2.: The interval lenghts in 1. attain values which asymptotically approach a periodicity of $s(n)$ in (multiple of $ \pi $)s.
Observing the numerical results, one finds that local mimima of $s(n)$ appear at values $n_i$ which have differences $d_i = n_i - n_{i-1}$ to the previous minimum of exactly either 104348 or 103993, here are the first and last values in the investigated range:
$n_1 = 73898, s(n_1) = 0.409953$
$n_2 = 178246, d_2 = 104348, s(n_2) = 0.409955$
$n_3 = 282594, d_3 = 104348, s(n_3) = 0.409954$
$n_4 = 386587, d_4 = 103993, s(n_4) = 0.409954$
$n_5 = 490935, d_5 = 104348, s(n_5) = 0.409956$
...
$n_{4795} = 499,696,605, d_{4795} = 104348, s(n_{4795}) = 0.410350$
$n_{4796} = 499,800,598, d_{4796} = 103993, s(n_{4796}) = 0.410356$
$n_{4797} = 499,904,946, d_{4797} = 104348, s(n_{4797}) = 0.410354 $
These difference values obey the properties:
$103993 - (33102 \cdot \pi) = -1.9129 \cdot 10^{-5}$
and
$104348 - (33215 \cdot \pi) = 1.1015 \cdot 10^{5}$
Hence there exist intervals which almost exactly reproduce the same minimum value of $s(n_i)$, and where the interval length is just very slightly more (less) than a multiple of $\pi$.
One further observes that the ratio of the number of minima with distance 104348 to the number of minima with distance 103993 equals (with very small deviation) the ratio of the deviations from the multiples of $\pi$, i.e. $1.9129 / 1.1015$ . For the observed 4797 minima, we have 4796 differences, where 3044 have the larger and 1752 the smaller value.
This means that, on the scale of the two observed "period lengths" between the minima, the occurence numbers of these two periods are balanced in such a way that an overall periodicity of $s(n)$ in (multiple of $ \pi $)s is generated. For short, the phase shifts - away from a periodicity in (multiple of $\pi$)s - cancel out. This periodicity guarantees that for arbitrarily large $n$, the conjecture will hold.
To illustrate this, let $n$ be arbitrary and let $n^*$ be the highest number for a minimum, for which $n^* \leq n$. Then
$$ s(n) = \sum_{k=0}^n|\cos k| - \frac{2}{\pi}n = \sum_{k=0}^{n^*}|\cos k| - \frac{2}{\pi} n^* + \sum_{k=n^* +1}^n|\cos k| - \frac{2}{\pi}(n-n^*) \\ = s(n^*) + \sum_{k=n^* +1}^n|\cos k| - \frac{2}{\pi}(n-n^*) $$
Shifting the index from $k$ to $m = k - (n^* - n_1)$, we have $$ s(n) = s(n_1) + (s(n^*) - s(n_1)) + \sum_{m=n_1 +1}^{n - (n^* - n_1)}|\cos (m + (n^* - n_1))| - \frac{2}{\pi}(n - n^*) $$
We have, with very small $\epsilon$, $(s(n^*) - s(n_1)) = \epsilon$. Likewise, with very small $\mu$, $|\cos (m + (n^* - n_1))| = |\cos (m + \mu)|$ - this merely illustrates the periodicity in (multiple of $\pi$)s and, more importantly, the fact that the sum over these terms is periodic over the large intervals.
Hence, with very small $\delta > 0$,
$$ s(n) = s(n_1) + \epsilon + \sum_{m=n_1 +1}^{n - (n^* - n_1)}|\cos (m + \mu)| - \frac{2}{\pi}(n - n^*) \\ = s(n - (n^* - n_1)) \pm \delta > 0.4 - \delta > 0 $$