For non zero positive reals a1,a2⋅⋅⋅an how to find
limx→∞(a1/x1+a1/x2+…+a1/xnn)nx?
It becomes indeterminant form 1∞. But difficult to solve by L'Hospital's Rule. By using A.M.-G.M. inequality it comes that limit is ≥a1a2⋅⋅⋅an. I also tried by using Squeeze. Please help.Thanks.
Answer
tl;dr
For a1,…,an positive numbers, and x>0, limx→∞(a1x1+a1x2+⋯+an1xn)nx=n∏k=1ak
Proof. Detailed approach, where we use the (low-order) Taylor expansions eu=1+u+o(u) and ln(1+u)=u+o(u) when u→0.
You can rewrite
(a1x1+a1x2+⋯+an1xn)nx=exp(nxlna1x1+a1x2+⋯+an1xn)=exp(nxln1nn∑k=1a1xk)=exp(nxln1nn∑k=1e1xlnak)
Since n is a constant and 1xlnak→x→∞0 for each k, we have
e1xlnak=1+1xlnak+o(1x)
for each k as x→∞, and therefore
(a1x1+a1x2+⋯+an1xn)nx=exp(nxln1nn∑k=1e1xlnak)=exp(nxln(1+1nxn∑k=1lnak+o(1x)))=exp(nx(1nxn∑k=1lnak+o(1x)))=exp(n∑k=1lnak+o(1))=exp(lnn∏k=1ak+o(1))=eo(1)n∏k=1ak→x→∞n∏k=1ak.
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