a1,a2,.am为m个正常数,证明lim（a1^n+a2^n+,.+am^n）开n方=max{a1,a2,.am}n趋向无穷

Question

a1,a2,.am为m个正常数,证明
lim（a1^n+a2^n+,.+am^n）开n方=max{a1,a2,.am}
n趋向无穷

Answer 1

不妨设am=max{a1,a2,.am} ,则
设A = lim n→∞ [（a1^n+a2^n+.+am^n）开n方]/am
A = lim n→∞ [（a1/am)^n+(a2/am)^n+.+(am/am)^n]开n方
ln A = lim n→∞ ln[（a1/am)^n+(a2/am)^n+.+(am/am)^n]/n
分子递减有下界为1,于是右边的极限为0,因此A=1,也即
lim n→∞ [（a1^n+a2^n+.+am^n）开n方] = am
原命题得证.