bn = (a1n + a2n+ a3n)1/n→max(a1,a2,a3)
Bew: oBdA: a3=max(a1,a2,a3)
\( \lim\limits_{n\to\infty} \) (a1n + a2n+ a3n)1/n
=\( \lim\limits_{n\to\infty} \) [a3n(a1n/a3n + a2n/a3n+ 1)]1/n
=\( \lim\limits_{n\to\infty} \) [ a3n*1/n ( (a1/a3)n + (a2/a3)n+ 1)1/n ]
=a3 \( \lim\limits_{n\to\infty} \) ( (a1/a3)n + (a2/a3)n+ 1)1/n
≤1 ≤1 ≤1
≤3 also: \( \lim\limits_{n\to\infty} \) [ ( (a1/a3)n + (a2/a3)n+ 1)1/n ] = 1
n-te Wurzel aus Zahl ∈]0,3] → 1
= a3
=max(a1,a2,a3)
Interessante Aufgabe!