Prove that:
limn→∞n√a1n+a2n+...+akn=max{a1,a2...ak}
I am familiar with the theorem which says that if
limn→∞anan−1=L
then,
limn→∞n√an=L
So, I tried evaluating the expreesion:
a1n+a2n+...+akna1n−1+a2n−1+...+akn−1
but pretty much got stuck here. Is this the right path to go?
Answer
n√anmax≤n√an1+⋯+ank≤n√k⋅anmax
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