<s>[INST] On the limit of $\sin^2 (\pi\sqrt{n^2+n})$ What is the limit of the sequence $\sin^2 (\pi\sqrt{n^2+n})$ as $n$ tends to infinity? 
My Attemp: I replace the square root with $n+\frac 12$ (its equivalent) and the rest is routine: 
$\lim \sin^2 (\pi\sqrt{n^2+n}) = \lim \sin^2 (\pi n + \frac{\pi}2) = 1$ 
Is this correct? 
 [/INST] Hint. The sequence is convergent.
As $n$ tends to $+\infty$, we may write
$$
\begin{align}
u_n &:=\sin^2 \left( \pi \sqrt{n^2+n }\right)\\
&=\sin^2 \left( \pi n \:\sqrt{1+\frac{1}{n}}\right)\\
&=\sin^2 \left( \pi n \:\left(1+\frac{1}{2n}+\mathcal{O}\left(\frac{1}{n^2}\right)\right)\right)\\
&=\sin^2 \left( \pi n +\frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right)\right)\\
&=\left((-1)^n\sin \left(\frac{\pi}{2}+\mathcal{O}\left(\frac{1}{n}\right)\right)\right)^2
\end{align}
$$
Can you finish it?
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