# TU Wien:Analysis UE (diverse)/Übungen SS19/Beispiel 40

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Man untersuche die Folge $\left\langle a_{n}\right\rangle _{n\in\mathbb{N}}$ auf Konvergenz und bestimme gegebenenfalls den Grenzwert.
$a_{n}=\frac{\sqrt{n+2}-\sqrt{n}}{\sqrt[3]{\frac{1}{n}}}$
\begin{align} 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{align}
\begin{align} \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{align}