高中數學題_級數問題

設f(x)= 1/x^(1/3)>0, x=1~10^6 (設 n=10^6)
f'(x)= (-1/3)x^(-4/3) <0, f"(x)=(4/9)x^(-7/3) >0 ,
圖形遞減凹向上,求積分∫[1~n] f(x) dx時, 梯形法所得>積分值, 則
Σ[k=1~n] 1/k^(1/3) - f(1)/2 - f(n)/2 > ∫[1~n] 1/x^(1/3) dx= (3/2)[ n^(2/3) -1]
Σ[k=1~n] 1/k^(1/3) > (1+ 0.01)/2 + (3/2)[ 10000-1]= 14999.005 ---(A)

又 1/k^(1/3) <∫[k-1, k] 1/x^(1/3) dx
則 Σ[k=1~n] 1/k^(1/3) < ∫[0~n] 1/x^(1/3) dx= (3/2)*n^(2/3)=15000 ---(B)

由(A),(B)得
14999.005 < Σ[k=1~10^6] 1/k^(1/3) < 15000

故所求之x的高斯值= 14999

贊同樓上0.0...

高中解這樣的題目, 會不會太難?

如果我沒計算錯, 答案應是 14999.
但這用到積分, 還用到convex function 的性質.