Proving $\lim\limits_{n→+∞}\sin(2\pi \sqrt{n^2+\sqrt{n}})=0$

Question

Prove that:

$$\underset{n\rightarrow +\infty }{\overset{}{\lim }} \ \sin\left( 2\pi \sqrt{n^{2} +\sqrt{n}}\right) =0$$

I thought to write the sequence like $f(n)g(n)$ with $f(n)$ infinitesimal and $g(n)$ bounded but I don't know how.

It is an exercise on first chapters of calculus textbook. I think it is possible to solve without integral or others advanced methods.

Answer 1

Since $\sin$ is periodic with period $2\pi$, you know that\begin{align}(\forall n\in\mathbb N):\sin\left(2\pi\sqrt{n^2+\sqrt n}\right)&=\sin\left(2\pi\sqrt{n^2+\sqrt n}-2\pi n\right)\\&=\sin\left(2\pi\left(\sqrt{n^2+\sqrt n}-n\right)\right).\end{align}But$$\lim_{n\to\infty}\sqrt{n^2+\sqrt n}-n=0.$$

Answer 2

Hint:

$n < \sqrt{n^2+\sqrt{n}} < n+\frac{1}{\log(n)}$ if $n>1$

Then, since $\sin(x)$ is crescent in $[2\pi n, 2\pi n + \frac{\pi}{2}]$

$0 = \lim_{n\to \infty} \sin(2\pi n) < \lim_{n\to \infty} \sin\left(2\pi\sqrt{n^2+\sqrt{n}}\right) < \lim_{n\to \infty} \sin\left(2\pi n+\frac{2\pi}{\log(n)}\right)$

Answer 3

I would write $$\sin \left( 2\pi \sqrt{n^2+ \sqrt{n}}\right) = \sin \left( 2\pi n \sqrt{1 + \frac{\sqrt{n}}{n^2}} \right) = \sin \left( 2\pi n \left( 1 + \frac{\sqrt{n}}{2n^2} + o\left( n^{-3/2}\right)\right)\right)$$ $$= \sin \left(\pi \frac{\sqrt{n}}{n} + o \left( n^{-1/2}\right) \right)$$

The argument of the sinus tends to $0$, and therefore, the whole sequence also.