Mittlerer Binomialkoeffizient/Anteil/Konvergenz/Aufgabe/Lösung

Es ist

${}{\begin{aligned}{\frac {x_{n+2}}{x_{n}}}&={\frac {{\binom {n+2}{{\frac {n}{2}}+1}}\cdot 2^{-n-2}}{{\binom {n}{\frac {n}{2}}}\cdot 2^{-n}}}\\&={\frac {1}{4}}\cdot {\frac {\frac {(n+2)(n+1)n\cdots {\left({\frac {n}{2}}+3\right)}{\left({\frac {n}{2}}+2\right)}}{{\left({\frac {n}{2}}+1\right)}{\frac {n}{2}}{\left({\frac {n}{2}}-1\right)}\cdots 2\cdot 1}}{\frac {n(n-1)(n-2)\cdots {\left({\frac {n}{2}}+2\right)}{\left({\frac {n}{2}}+1\right)}}{{\frac {n}{2}}{\left({\frac {n}{2}}-1\right)}{\left({\frac {n}{2}}-2\right)}\cdots 2\cdot 1}}}\\&={\frac {1}{4}}\cdot {\frac {(n+2)(n+1)}{{\left({\frac {n}{2}}+1\right)}\cdot {\left({\frac {n}{2}}+1\right)}}}\\&={\frac {(n+2)(n+1)}{(n+2)\cdot (n+2)}}\\&={\frac {n+1}{n+2}}.\end{aligned}}$

Daher ist

${}x_{n+2}={\frac {n+1}{n+2}}x_{n}={\frac {n+1}{n+2}}\cdot {\frac {n-1}{n}}\cdot {\frac {n-3}{n-2}}\cdots {\frac {3}{4}}\cdot {\frac {1}{2}}x_{0}\,$

und die Nullkonvergenz folgt aus

Aufgabe.

Zur gelösten Aufgabe