[math] \newcommand{\smallfrac}[2]{\frac{#1}{#2}} \newcommand{\medfrac}[2]{\frac{#1}{#2}} \newcommand{\textfrac}[2]{\frac{#1}{#2}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\e}{\operatorname{e}} \newcommand{\B}{\operatorname{B}} \newcommand{\Bbar}{\overline{\operatorname{B}}} \newcommand{\pr}{\operatorname{pr}} \newcommand{\dd}{\operatorname{d}\hspace{-1pt}} \newcommand{\E}{\operatorname{E}} \newcommand{\V}{\operatorname{V}} \newcommand{\Cov}{\operatorname{Cov}} \newcommand{\Bigsum}[2]{\mathop{\textstyle\sum}_{#1}^{#2}} \newcommand{\ran}{\operatorname{ran}} \newcommand{\card}{\#} \newcommand{\mathds}{\mathbb}[/math]

Let [math]\Bbar_1(0)[/math] denote the unit ball of [math]\mathbb{R}^d[/math].

• Use Lemma to compute [math]\lambda^d(\Bbar_1(0))[/math] for [math]d=1,\dots,10[/math].
• Compute [math]\lim_{d\rightarrow\infty}\lambda^d(\Bbar_1(0))[/math].
• Show that [math]\lambda^d(\Bbar_1(0))=\medfrac{\pi^{d/2}}{\Gamma(d/2+1)}[/math] holds, where [math]\Gamma\colon(0,\infty)\rightarrow\mathbb{R}[/math] denotes the Gamma function.