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? 
 A: 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?
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\sin^{2}\pars{\pi\root{n^{2} + n}} & =
\sin^{2}\pars{\pi n\root{1 + {1 \over n}}} =
\sin^{2}\pars{\pi n + \pi n\bracks{\root{1 + {1 \over n}} - 1}}
\\[5mm] & =
\bracks{\pars{-1}^{n}\sin\pars{\pi n\bracks{\root{1 + {1 \over n}} - 1}}}^{2}
\\[5mm] & =
\sin^{2}\pars{\pi \over \root{1 + 1/n} + 1}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, \sin^{2}\pars{\pi \over 2} = \bbx{\ds{1}}
\end{align}
