Divergence of the sequence $a_n= \frac{n^{2} + 2n + 35}{n^2 + n + 1} \sin(n^{2}) $

Question

Are there any methods to prove that the sequence $\left\{a_{n} = \dfrac{n^{2} + 2n + 35}{n^2 + n + 1} \sin(n^{2})\right\}$ is divergent?

To be honest, I'm not good at solving any problems relating to trigonometry. So proving the divergence of this trigonometric sequence above is quite hard. I've tried to use contradiction that means suppose the sequence is convergent and then all of its subsequences are convergent, too. Next I try to find one subsequence of $a_{n}$ is divergent and I get contradiction. But this is not easy to find a subsequence like that.

Thank you for your help.

Answer 1

For clarity I write out why the proposed duplicate answers the question. Note that

$$ a_n =\frac{n^2+2n+35}{n^2+n+1}\sin(n^2) = \sin(n^2) + \frac{(n+34)\sin(n^2)}{n^2+n+1}$$

Now $$\left|\frac{(n+34)\sin(n^2)}{n^2+n+1}\right| \le \frac{35n}{n^2} \to 0$$ and by the equidistribution of $n^2 \pmod {2\pi}$ as explained in the linked posts e.g. Convergence of $\sum_n \frac{|\sin(n^2)|}{n}$ or $\sin (n^2)$ diverges , it follows that there are infinitely many $n$ such that $\sin(n^2)>1/2$, and there are infinitely many $n$ such that $\sin(n^2)<-1/2$. It follows that there are infinitely many $n$ such that $a_n>1/4$, and infinitely many $n$ such that $a_n < -1/4$. Hence, $a_n$ has no limit.