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令a_n = n^(1/n)-1>=0 那麼n>1時n =(1+a_n)^n = 1+n*a_n+n(n-1)/2*a_n^2+…+a_n^n < 1+n(n-1)/2*a_n^2 所以0 0,即n^(1/n)->1
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