Prove that a sequence is convergent and determine its limit 
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?
 A: 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?
A: 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 $\lim_{n\to\infty} \cos^2 \frac{-\pi}{2n+4}=1$.
A: 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 $x_n\to\cos^20=1$.
