Aloha :)
$$\sin\left(\frac{n\pi}{2n+1}\right)=\sin\left(\frac{\pink{\frac1n}\cdot n\pi}{\pink{\frac1n}\cdot(2n+1)}\right)=\sin\left(\frac{\pi}{2+\frac1n}\right)\to\sin\left(\frac{\pi}{2+0}\right)=1$$
$$\sqrt{n^2+n+1}-\sqrt{n^2+1}=\frac{(\overbrace{\sqrt{n^2+n+1}}^a-\overbrace{\sqrt{n^2+1}}^b)\pink{(\overbrace{\sqrt{n^2+n+1}}^a+\overbrace{\sqrt{n^2+1}}^b)}}{\pink{(\sqrt{n^2+n+1}+\sqrt{n^2+1})}}$$$$\qquad=\frac{\overbrace{(n^2+n+1)}^{a^2}-\overbrace{(n^2+1)}^{b^2}}{n\sqrt{1+\frac1n+\frac{1}{n^2}}+n\sqrt{1+\frac{1}{n^2}}}=\frac{\green n}{\green n\sqrt{1+\frac1n+\frac{1}{n^2}}+\green n\sqrt{1+\frac{1}{n^2}}}$$$$\qquad=\frac{1}{\sqrt{1+\frac1n+\frac{1}{n^2}}+\sqrt{1+\frac{1}{n^2}}}\to\frac{1}{\sqrt{1+0+0}+\sqrt{1+0}}=\frac12$$
$$\log(n+1)-2\log(\sqrt n)=\log(n+1)-\log((\sqrt n)^2)=\log(n+1)-\log(n)$$$$\qquad=\log\left(\frac{n+1}{n}\right)=\log\left(1+\frac1n\right)\to\log(1+0)=0$$
Die Folge \((x_n)\) konvergiert also gegen \((1\big|\frac12\big|0)\).
