Limit of the sequence $\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$ I would like to calculate the following limit: ${\lim _{n \to \infty }}\sin \left( {2\pi \sqrt {{n^2} + n} } \right)$
I am not sure if this limit exists...
 A: Hint: Note that
$$\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)=\frac{\left(\sqrt{n^2+n}-\left(n+\frac{1}{2}\right)\right)\left(\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)\right)}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}=\frac{-1/4}{\sqrt{n^2+n}+\left(n+\frac{1}{2}\right)}.$$
Thus
$$\sin\left(2\pi\sqrt{n^2+n}\right)=\sin\left(2\pi n+\pi - \frac{\pi/2}{\sqrt{n^2+n}+\left(n+\frac{1}{2}  \right)}\right).$$
A: Assuming that $n$ is an integer variable,
$$\eqalign{
  \sin(2\pi\sqrt{n^2+n})
  &=\sin\bigl(2\pi\sqrt{n^2+n}-2\pi n\bigl)\cr
  &=\sin\left(\frac{2\pi}{\sqrt{1+\frac1n}+1}\right)\cr
  &\to\sin\pi\cr
  &=0\ .\cr}$$
If on the other hand $n$ is a real variable, then as $n$ tends to $\infty$, the expression $\sqrt{n^2+n}$ takes all positive real values, so $\sin(2\pi\sqrt{n^2+n})$ keeps on oscillating between $1$ and $-1$, and has no limit.
A: Hint: $2\pi \sqrt {n^2 +n} = 2\pi n\sqrt {1 +1/n}.$
A: With a limited development :
$$\eqalign{
  \sin(2\pi\sqrt{n^2+n})
  &=\sin(2\pi n\sqrt{1+\frac1n})\cr
  &=_{n\rightarrow\infty}\sin(2\pi n\times(1+\frac1{2n}+o(\frac1n)))\cr
  &=\sin(2\pi \times(n+\frac12+o(1)))\cr
  &=\sin(\pi + 2\pi n + o(1))\cr
  &=-\sin(2\pi n + o(1)) \longrightarrow 0\ \cr}$$
A: Notice, let $n$ be any integer then we have $$\lim_{n\to \infty}\sin(2\pi\sqrt{n^2+n})$$ $$=\lim_{n\to \infty}\sin(-2n\pi+2\pi\sqrt{n^2+n})$$  $$=\lim_{n\to \infty}\sin\left(2\pi n\sqrt{1+\frac{1}{n}}-2\pi n\right)$$ $$=\lim_{n\to \infty}\sin\left(2\pi n \left(1+\frac{1}{n}\right)^{1/2}-2\pi n\right)$$ Since, $n$ is large enough hence using the binomial expansion of $\left(1+\frac{1}{n}\right)^{1/2}$ $$=\lim_{n\to \infty}\sin\left(2\pi n \left(1+\frac{1}{2}\left(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)\right)-2\pi n\right)$$  $$=\lim_{n\to \infty}\sin\left(2\pi n \left(\frac{1}{2}\left(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)\right)\right)$$  $$=\lim_{n\to \infty}\sin\left(\pi+(2\pi) O\left(\frac{1}{n}\right)\right)=\sin \pi=0$$ But, if $n$ is a real number then $\sin(2\pi\sqrt{n^2+n})$ can have any value within domain $[-1, 1]$ hence it has no definite limit (i.e. unique value) 
