Man untersuche die Folge ⟨an⟩n∈N{\displaystyle \left\langle a_{n}\right\rangle _{n\in \mathbb {N} }} auf Konvergenz und bestimme gegebenenfalls den Grenzwert.
an=n+2−n1n3=n+2−n1n3⋅n+2+nn+2+n=n+2−n1n3⋅(n+2+n)=21n3⋅(n⋅(1+2n)+n)=21n3⋅n⋅(1+2n+1)=2n−13⋅n12⋅(1+2n+1)=2n12−13⋅(1+2n+1)=2n16⋅(1+2n+1)=2n6⋅(1+2n+1){\displaystyle {\begin{aligned}a_{n}&={\frac {{\sqrt {n+2}}-{\sqrt {n}}}{\sqrt[{3}]{\frac {1}{n}}}}\\&={\frac {{\sqrt {n+2}}-{\sqrt {n}}}{\sqrt[{3}]{\frac {1}{n}}}}\cdot {\frac {{\sqrt {n+2}}+{\sqrt {n}}}{{\sqrt {n+2}}+{\sqrt {n}}}}\\&={\frac {n+2-n}{{\sqrt[{3}]{\frac {1}{n}}}\cdot \left({\sqrt {n+2}}+{\sqrt {n}}\right)}}\\&={\frac {2}{{\sqrt[{3}]{\frac {1}{n}}}\cdot \left({\sqrt {n\cdot \left(1+{\frac {2}{n}}\right)}}+{\sqrt {n}}\right)}}\\&={\frac {2}{{\sqrt[{3}]{\frac {1}{n}}}\cdot {\sqrt {n}}\cdot \left({\sqrt {1+{\frac {2}{n}}}}+1\right)}}\\&={\frac {2}{n^{-{\frac {1}{3}}}\cdot n^{\frac {1}{2}}\cdot \left({\sqrt {1+{\frac {2}{n}}}}+1\right)}}\\&={\frac {2}{n^{{\frac {1}{2}}-{\frac {1}{3}}}\cdot \left({\sqrt {1+{\frac {2}{n}}}}+1\right)}}\\&={\frac {2}{n^{\frac {1}{6}}\cdot \left({\sqrt {1+{\frac {2}{n}}}}+1\right)}}\\&={\frac {2}{{\sqrt[{6}]{n}}\cdot \left({\sqrt {1+{\frac {2}{n}}}}+1\right)}}\end{aligned}}}
limn→∞an=limn→∞2n6⋅(1+2n+1)=limn→∞2n6⏟∞⋅(1+2n+1)⏟2=2∞⋅2=0{\displaystyle {\begin{aligned}\lim _{n\to \infty }a_{n}&=\lim _{n\to \infty }{\frac {2}{{\sqrt[{6}]{n}}\cdot \left({\sqrt {1+{\frac {2}{n}}}}+1\right)}}\\&=\lim _{n\to \infty }{\frac {2}{\underbrace {\sqrt[{6}]{n}} _{\infty }\cdot \underbrace {\left({\sqrt {1+{\frac {2}{n}}}}+1\right)} _{2}}}\\&={\frac {2}{\infty \cdot 2}}\\&=0\end{aligned}}}