Question

用比较判别法的极限形式判别∑(n+1)/(n^2+n+1)的敛散性

Answer 1

lim(n->∞)【(n+1)/(n^2+n+1)】/(1/n)
=lim(n->∞) n(n+1)/(n^2+n+1)
=1
∑(n+1)/(n^2+n+1)和 ∑1/n 一样发散

Answer 2

①∑(n+1)/(n^2+n+1)
<∑(n+1)/(n^2+n)
=∑(1/n)
因为调和级数∑(1/n)发散
②∑(n+1)/(n^2+n+1)
>∑(n+1)/(n^2+2n+1)
=∑(1/(n+1))
因为调和级数∑(1/(n+1))发散
由比较判别法得∑(n+1)/(n^2+n+1)发散