Does $n^2-\sin(n)\sqrt{n}$ go to $+\infty$ as $n \to +\infty$?

Question

How does one prove that $n^2-\sin(n)\sqrt{n}$ converges to infinity?

I've attempted it below, but I'm not too confident.

I say for any $M>0$ take $N= \lceil (M+1)^2 \rceil+1$, then $n>(M+1)^2$ gives $n^2 > (M+1)^4$ and $\sqrt{n} > M+1$. Also, $\sin(n) \le 1$ therefore $\sin(n)\sqrt{n} \le \sqrt{n}$ hence $-\sin(n)\sqrt{n} \ge -\sqrt{n}$ hence:

$$n^2-\sin(n)\sqrt{n} \ge n^2-\sqrt{n} > M. $$

Therefore it converges for $+\infty$.

Answer 1

You are almost correct: if $n>M>2$ then $$n^2-\sin(n)\sqrt{n}\geq n^2-\sqrt{n}>n^2-n=n(n-1)>M.$$ You may also say that $$n^2-\sin(n)\sqrt{n}=n^2\left(1-\frac{\sin(n)}{n\sqrt{n}}\right)$$ which goes to $+\infty$ because $n^2\to +\infty$ and $\frac{\sin(n)}{n\sqrt{n}}\to 0$ because $$0\leq \frac{|\sin(n)|}{n\sqrt{n}}\leq \frac{1}{n}.$$

Answer 2

No, it is not correct. From $n^2>(M+1)^4$ and $\sqrt n>M+1$, there's nothing that you can deduce about $n^2-\sqrt n$.

Note that $n^2-\sqrt n=\sqrt n(n^{3/2}-1)>\sqrt n$. Use this to prove that your limit is $+\infty$.

Answer 3

It is not clear why $n^{2}-\sqrt{n}>M$ at the end. Perhaps one does in the following way: We see that $\sqrt{n}<\dfrac{1}{2}n^{2}$ for $n\geq 2$ (by taking square both sides, then $n<(1/4)n^{4}$, or $n^{4}>4n$ can be done by induction easily for $n\geq 2$), then $n^{2}-\sqrt{n}>(1/2)n^{2}>M$ for all $n\geq \sqrt{2N}$, of course, one lets $N>0$ such that $\sqrt{2N}>2$, for example, $2N>4$, that is, $N>2$.