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令 fn代表原级限值。
lnfn
= lim{x->oo} (1/n)[ln(n+1)+ln(n+2)+...+ln(2n)] - (nlnn)/n
= lim{x->oo} (1/n)[ln(1+1/n)+ln(1+2/n)+...+ln(1+n/n)]
= ∫[0,1] ln(1+x)dx, where 1/n = dx, i/n = x, 用定积分求极限
= xln(1+x) - ∫[0,1] x/(1+x)dx, integration by parts
= (1+x)ln(1+x) - x|[0,1]
= 2ln2-1
Therefore, fn = e^(2ln2-1)
lnfn
= lim{x->oo} (1/n)[ln(n+1)+ln(n+2)+...+ln(2n)] - (nlnn)/n
= lim{x->oo} (1/n)[ln(1+1/n)+ln(1+2/n)+...+ln(1+n/n)]
= ∫[0,1] ln(1+x)dx, where 1/n = dx, i/n = x, 用定积分求极限
= xln(1+x) - ∫[0,1] x/(1+x)dx, integration by parts
= (1+x)ln(1+x) - x|[0,1]
= 2ln2-1
Therefore, fn = e^(2ln2-1)
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