How can I find the limit of the following sequence: $$\sin ^2 (\pi \sqrt{n^2 + n})$$
I feel that I will use the identity $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n})), $$
But then what? how can I deal with the limit of $\cos (2 \pi \sqrt{n^2 + n})$? I know that $\cos (n\pi) = (-1)^n$, if $n$ is a positive integer but then what?
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You can check $\sin^2(\pi\sqrt{n^2 + n})=\sin^2(\pi\sqrt{n^2 + n}-\pi n)$. So $$\sin^2(\pi\sqrt{n^2 + n})=\sin^2 \pi\frac{n}{\sqrt{n^2 + n}+n}\to \sin^2\frac{\pi}{2}=1.$$
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Assuming that you want to know more than the limit itself.
Using $$\sin ^2 (\pi \sqrt{n^2 + n}) = \frac{1}{2}(1- \cos(2 \pi \sqrt{n^2 + n}))$$ use Taylor series $$\sqrt{n^2 + n}=n+\frac{1}{2}-\frac{1}{8 n}+O\left(\frac{1}{n^2}\right)$$ $$2\pi\sqrt{n^2 + n}=(2n+1)\pi -\frac{\pi}{4 n}+O\left(\frac{1}{n^2}\right)$$ $$\cos(2 \pi \sqrt{n^2 + n})\sim -\cos \left(\frac{\pi }{4 n}\right)=-1+\frac{\pi ^2}{32 n^2}+O\left(\frac{1}{n^4}\right)$$ $$\sin ^2 (\pi \sqrt{n^2 + n}) =1-\frac{\pi ^2}{64 n^2}+O\left(\frac{1}{n^4}\right)$$ then the limit and how it is approached.
Using $n=1000$, the "exact" value would be $$\sin ^2\left(10 \sqrt{10010} \pi \right)\approx 0.99999984594$$ while the above approximation would give $$1-\frac{\pi ^2}{64000000}\approx 0.99999984578$$
Intuitively, $n^2+n$ is almost exactly halfway between $n^2$ and $(n+1)^2$, and so $\sqrt{n^2+n}$ is very close to a half-integer. This means $\sin(\pi\sqrt{n^2+n})$ should be very close to $\pm 1$, and in either case $\sin^2(\pi\sqrt{n^2+n})$ will be very nearly $1$.
More formally, we have $$\sin^2(\pi\sqrt{n^2+n})=\sin^2[\pi(\sqrt{n^2+n}-n)] $$ Now, multiplying by $1$ in the form $\frac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n}$ gives
$$ \sin^2[\pi(\sqrt{n^2+n}-n)]=\sin^2\left(\pi\frac{n}{\sqrt{n^2+n}+n}\right) $$
Since $\lim_{n \to \infty} \frac{n}{\sqrt{n^2+n}+n}=\frac{1}{2}$ and $\sin^2$ is continuous, this sequence has limit $\sin^2 \frac{\pi}{2}=1$.