Show that $\sin n^2$ is divergent.

Question

Problem

Show that the number sequence $\sin n^2$ is divergent.

My Proof

Assume that $\lim\limits_{n \to \infty}\sin n^2=L.$ Then,

\begin{align*} \int_0^{+\infty}\cos x {\rm d}x&=\lim_{n \to \infty}\int_0^{n^2}\cos x {\rm d}x\\ &=\lim_{n \to \infty}\sum_{k=0}^{n-1}\int_{k^2}^{(k+1)^2}\cos x {\rm d}x\\ &=\lim_{n \to \infty}\sum_{k=0}^{n-1}\left[\sin(k+1)^2-\sin k^2\right]\\ &=\lim_{n \to \infty}\sin n^2\\ &=L. \end{align*}

But, in fact, the infinite integral on the left side is divergent, since $$ \int_0^{+\infty}\cos x {\rm d}x=\lim_{t \to +\infty}\int_0^t\cos x {\rm d}x=\lim_{t \to +\infty}\sin t,$$ the last limit does not exist. Thus, the contradiction comes out.

Please correct me if I'm wrong.Thanks in advance.

Answer 1

No, it is not correct. It is true that$$\int_0^{+\infty}\cos x\,\mathrm dx=\lim_{n \to \infty}\int_0^{n^2}\cos x\,\mathrm dx,$$but with $n\in\mathbb{R}^+$. You are assuming that $n\in\mathbb N$.

Answer 2

Your proof does not make any sense, both for the fact that $\int_{0}^{+\infty}\cos(x)\,dx$ is meaningless and for the fact that the primitive of $\cos(x)$ has very little to do with the behaviour of $\sin(x^2)$ over the integers.

What we can say is that $\lim_{n\to +\infty}\sin(n^2)=L$ implies $L=0$, since the series $$ \sum_{n\geq 1}\frac{\sin n^2}{n} $$ is convergent by Weyl's inequality. In order to actually show that $\lim_{n\to +\infty} \sin(n^2)$ does not exist, it is enough to prove that the sequence $\{n^2\pmod{\pi}\}_{n\geq 1}$ has an accumulation point which differs from zero. This follows from the fact that the forward difference $(n+1)^2-n^2 = 2n+1$ is equidistributed $\pmod{\pi}$.