How find the limit $\lim_{n\to\infty}\frac{\sum_{k=1}^n|\cos(k^2)|}{n}$.

Question

Compute $$\displaystyle\lim_{n\to\infty}\dfrac{\displaystyle\sum_{k=1}^n|\cos(k^2)|}{n}$$.

I guess is $\dfrac{2}{\pi}$,because the summation is essentially equal to computing the average value of $|\cos k|$ on the interval from $[0, \pi]$, which is $\boxed{\dfrac{2}{\pi}}$,it's right?Thanks

It would be helpful that fractional part of $n^{2}\alpha$ for irrational $\alpha$ is equidistributed on $[0,1)$. — Seewoo Lee, Apr 12 '20 at 11:05
so the $\lim_{n\to+\infty}\dfrac{\sum_{k=1}^{n}|\cos{k^m}|}{n}=\dfrac{2}{\pi}?$,\forall $m\in N^+$? — math110, Apr 12 '20 at 13:05

score 3 · Accepted Answer · answered Apr 13 '20 at 02:10

Yes. the limit is $\frac{2}{\pi}$. Let $m$ be any integer $> 1$.

Since $\pi \not\in\mathbb{Q}$, equidistributed theorem tell us $\left\{\frac{k}{2\pi}\right\}$ is equidistributed on $[0,1]$.
Apply van der Corput's difference theorem $m-1$ times, we find $\left\{\frac{k^m}{2\pi}\right\}$ is also equidistributed on $[0,1]$.
Since $|\cos(k^m)| = \left|\cos\left(2\pi\left\{\frac{k^m}{2\pi}\right\}\right)\right|$, by Riemann integral criterion for equidistribution, we have

$$\lim\limits_{n\to\infty}\frac1n\sum_{k=1}^n |\cos(k^m)| = \int_0^1 |\cos(2\pi x)| dx = \frac2{\pi}$$

How find the limit $\lim_{n\to\infty}\frac{\sum_{k=1}^n|\cos(k^2)|}{n}$.

1 Answers1