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I have to prove that the sequence $(x_n)_{n \in \mathbb R}$ defined by: $x_n=\cos^2(\pi \sqrt{(n+1)(n+3})$ is convergent and determine its limit.

If I find that the limit exists (and its finite) doesn' t that mean that the sequence is convergent?

I wrote $x_n=1-\sin^2(\pi \sqrt{(n+1)(n+3})$ and for $n\rightarrow \infty$, $\sin^2(\pi \sqrt{(n+1)(n+3}) \sim \sin^2(\pi \sqrt{n^2})=\sin^2(\pi n)=0$. So the limit is actually $1$. Is this correct?

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    How do you arrive at the $\sim$? You cannot conclude it immediately from $\sqrt{(n+1)(n+3)}\sim n$ – 2017-02-23

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we have $$\cos(\pi\sqrt{n^2+4n+3}-\pi n+\pi n)=$$ $$\cos\left(\frac{\pi(4n+3)}{\sqrt{n^2+4n+3}+n}+\pi n\right)$$= $$\cos\left(\frac{\pi\left(4+\frac{3}{n}\right)}{\sqrt{1+\frac{4}{n}+\frac{3}{n^2}}}+\pi n\right)$$ =$$\cos\left(\frac{\pi \left(4+\frac{3}{n}\right)}{\sqrt{1+\frac{4}{n}+\frac{3}{n^2}}+1}\right)\cos\left(\pi n\right)$$ since $$\sin(\pi n)=0$$ can you finish this?

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    I like this answer a lot! Yes, its pretty clear from this point onwards, with square raised and $n \rightarrow \infty$. – 2017-02-23
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    thank you since Math is my live! – 2017-02-23
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Note that $\sqrt{n^2+4n+3}=\left( n+2\right)\sqrt{1-\left( n+2\right)^{-2}}\approx n+2-\frac{1}{2n+4}$, so since $\cos^2 y$ has period $\pi$ you're seeking $\lim_{n\to\infty} \cos^2 \frac{-\pi}{2n+4}=1$.

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    Can you please explain where the $\approx$ relationship with the square root comes from? – 2017-02-23
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    @JohnZobias The binomial theorem for $(1-z)^{1/2}$. – 2017-02-23
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Note that $(n+1)(n+3)=(n+2)^2-1$. Also, $$\left((n+2)-\frac1{2(n+2)}\right)^2=(n+2)^2-1+\frac1{4(n+2)^2}$$ is slightly bigger than $(n+2)^2-1$, whereas $$\left((n+2)-\frac1{n+2}\right)^2=(n+2)^2-2+\frac1{(n+2)^2}$$ is strictly smaller than $(n+2)^2-1$ (because $\frac1{(n+2)^2}<1$). We conclude $$ (n+2)-\frac1{n+2}<\sqrt{(n+1)(n+3)}<(n+2)-\frac1{2(n+2)}$$ And as $\cos (\pi a)=\pm \cos(\pi(a-n-2))$, we conclude that $x_n=\cos^2 (\pi\delta_n)$ where $-\frac1{n+2}<\delta_n<-\frac1{2(n+2)}$. We see $\delta_n\to 0$ and hence $x_n\to\cos^20=1$.