Prove that $a_n=-\sqrt{n}\cos\sqrt{n}-\sin\sqrt{n}+\cos 1+\sin 1$ diverges Can I simply pass the limit in
$a_n=-\sqrt{n}\cos\sqrt{n}-\sin\sqrt{n}+\cos 1+\sin 1$ 
and say that this sequence diverge, because $\sqrt{n}\rightarrow\infty$?
I have found the manual solution of a book which has this exercise.

Take $\alpha=\frac{1}{2}$. Does the sequence $a_n=\int_1^{\infty}\sin x^{\alpha}\,dx$ converge or diverge? 

The development of its solution leads to my problem, but the manual gives only a hint. Here it goes.

Note that being $k_n$ the sequence $k_1=31$, $k_2=314$, $k_3=314$, $k_4=3141,\dots$, i.e, $k_n$ is $10^n$ times the approximation of $\pi$ with $n$ digits. Then 
  $$\lim\sqrt{k_n^2}\cos\sqrt{k_n^2}=+\infty \ \ \  (\mbox{Verify!})$$
The same way we can construct another sequence $p_n$ of natural number such that
$$\lim\sqrt{p_n^2}\cos\sqrt{p_n^2}=-\infty \ \ \  (\mbox{Think about it!})$$

Could anyone help me with this?
 A: We will show that the sequence defined by
$$n\cos n$$
is divergent. Now, suppose by way of contradiction that $\cos n$ converges to $0$. I present a slightly modified version of the proof here to show this is impossible. Define the sequences
$$(x_n,y_n)=(\cos n,\sin n)$$
Then
$$x_{n+1}=x_n\cos 1-y_n\sin 1$$
$$y_{n+1}=x_n \sin 1+y_n \cos 1$$
This implies
$$0=\lim_{n\to\infty}x_{n+1}=\lim_{n\to\infty}(x_n\cos 1-y_n\sin 1)=\cos 1\lim_{n\to\infty}x_n -\sin 1\lim_{n\to\infty}y_n=-\sin 1\lim_{n\to\infty}y_n$$
$$\Rightarrow \lim_{n\to\infty}y_n=0$$
However, this leads to
$$1=\lim_{n\to\infty} 1=\lim_{n\to\infty}(x_n^2+y_n^2)=\lim_{n\to\infty}x_n^2+\lim_{n\to\infty}y_n^2=0+0=0$$
which is a contradiction. Now, since $\cos n$ does not approach $0$, there exists some subsequence of the natural numbers (let us call it $b_k$) and constant $\epsilon>0$ such that $|\cos b_k|>\epsilon$ for all $k\in\mathbb{N}$. This implies that
$$|\sqrt{n}\cos\sqrt{n}|$$
evaluated at $b_k^2$ is
$$\left|\sqrt{b_k^2}\cos\sqrt{b_k^2}\right|=|b_k\cos b_k|>b_k\epsilon$$
which clearly goes to infinity as $k$ goes to infinity. Since every other term in the sequence is bounded, and $\left|\sqrt{b_k^2}\cos\sqrt{b_k^2}\right|$ has an unbounded magnitude, we conclude 
$$-\sqrt{n}\cos\sqrt{n}-\sin\sqrt{n}+\cos 1+\sin 1$$
diverges.
