题目错了吧,应是“ 1/(n³+2n²) ”吧
1/(n³+2n²) < 1/(n³+2n²-3n)
1/(n³+2n²-3n) = 1/[n(n+3)(n-1)]
= (1/2) [(n+3)+(n-1)-2n]/ [n(n+3)(n-1)]
= 1/[2n(n-1)] +1/[2n(n+3)] - 1/[(n+3)(n-1)]
= [1/(n-1) - 1/n]/2 + [1/n - 1/(n+3)]/6 - [1/(n-1) - 1/(n+3)]/4
因此,∑(n=1,∞) 1/(n³+2n²-3n) 是收敛的,
则∑(n=1,∞) 1/(n³+2n²)也是收敛的.