□投稿者/ soredeha 一般人(5回)-(2007/05/03(Thu) 08:11:12)
| グラフから ∫[1,n]logxdx<log1+・・+logn<∫[1,n+1]logxdx nlogn-(n-1)<log1+・・+logn<(n+1)log(n+1)-n logn-(1-1/n)<(log1+・・+logn)/n<(1+1/n)log(n+1)-1 -(1-1/n)<(log1+・・+logn)/n-logn<(1+1/n)log(n+1)-1-logn =log(n+1)-logn+(1/n)log(n+1)-1 =log(1+1/n)+{(n+1)/n}log[n+1]√(n+1)-1 → -1 ( n → ∞ )
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