设f(x)在区间[1,+∞)上单调减少且非负的连续函数一∫0nf(x)dx(n=1,2,…). (1)证明: (2)证明:反常积分∫1+∞f(x)dx与无穷级数同敛散.

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问题 设f(x)在区间[1,+∞)上单调减少且非负的连续函数一∫0nf(x)dx(n=1,2,…).
(1)证明:
(2)证明:反常积分∫1+∞f(x)dx与无穷级数同敛散.

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答案(1)由f(x)单调减少,故当k≤x≤k+1时, f(k+1)≤f(x)≤f(k). 两边从k到k+1积分,得 ∫kk+1f(k+1)dx≤∫kk+1f(x)dx≤∫kk+1f(k)dx,即 f(k+1)≤∫kk+1f(x)dx≤f(k). [*] 即{an}有下界.又 an+1一an=f(n+1)一∫nn+1f(x)dx≤0, 即数列{an}单调减少,所以[*]存在. (2)由于f(x)非负,所以∫1xf(t)dt为x的单调增加函数.当n≤x≤n+1时,∫1nf(t)dt≤∫1xf(t)dt≤∫1n+1f(t)dt,所以 [*]

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