Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

Question

I am reading "An Introduction to Analysis vol. 2" (in Japanese) by Kazuo Matsuzaka.

There is the following example in this book:

Let $f_n(x) := \frac{\sin n^2 x}{n}$ for $n = 1,2,\dots$.
Then $f(x):=\lim_{n\to\infty} f_n(x)=0$, so $f'(x) = 0$.
On the other hand, $f_n^{'}(x) = n \cos n^2 x$ and $f_n^{'}(x)$ doesn't converge as $n\to\infty$.
So, $f_n\to f$ and, $f_n$ and $f$ are differentiable, but $f_n^{'}\to f^{'}$ doesn't hold.

The author wrote $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.
But I cannot prove that $n\cos n^2x$ doesn't converge as $n\to\infty$ even for $x=1$.

Please prove that $n\cos n^2 x$ doesn't converge as $n\to\infty$ for any $x\in\mathbb{R}$.

Answer 1

Hint

For any large positive integer $N$, you should be able to find integers $r,s$ such that $r > N$ and $s > N$ and such that $xr^2$ is reasonably close to $\pi \pmod{2\pi}$, while $xs^2$ is reasonably close to $0 \pmod{2\pi}.$

Answer 2

I'm not sure how rigorous you need to be, but it's a necessary condition for this sequence to converge that $\cos(n^2 x) \longrightarrow 0$. This is impossible.

To see why, consider that if $k^2x \approx (2m +1) \pi$ and $(k+1)^2x \approx (2n +1) \pi$, then $(2k+1)x \approx 2j \pi$ for some integer $j$. It follows from this that $2jk^2 \approx (2k+1)(2n+1)$, but the left hand side is even, and the right hand side is odd. So, it's impossible.

Answer 3

The sequence $\{n \cos n^2 \pi\}_{n=0}^\infty$ equals $\{0, 1, -4, 9, -16, 25\dots\}$

Answer 4

As @user20672 said, if $\lim_{n\to\infty} n\cos n^2 x\in\mathbb{R}$ then $\lim_{n\to\infty}\cos n^2 x=0$.

Assume that $\lim_{n\to\infty} n\cos n^2 x=A\in\mathbb{R}$ for some $x\in\mathbb{R}$.
Let $a_n := \cos n^2x$.
Then $\lim_{n\to\infty} a_n =\lim_{n\to\infty} \frac{1}{n} \times n\cos n^2x= 0 \times A=0$.
Since $\{a_{2n}\}$ is a subsequence of $\{a_n\}$, so $\lim_{n\to\infty} a_{2n}=0$.

On the other hand, $a_{2n}=\cos(2n)^2x=\cos4n^2x=2(\cos2n^2x)^2-1=2(2(\cos n^2x)^2-1)^2-1=2(2a_n^2-1)^2-1$.
So, $\lim_{n\to\infty}a_{2n}=\lim_{n\to\infty}2(2a_n^2-1)^2-1=2(2\times0^2-1)^2-1=2-1=1$.

This is a contradiction. So $\lim_{n\to\infty} n\cos n^2 x$ diverges.