高等数学 求级数的敛散性 ∑2n+1分之n+1 n趋于∞

2025-05-21 11:04:46
推荐回答(1个)
回答1:

∑(n:0->∞) (n+1)/(2n+1)

(n+1)/(2n+2) < (n+1)/(2n+1)
1/2 < (n+1)/(2n+1) (1)

(n+1)/(2n+1) < (2n+1)/(2n+1) =1 (2)

1/2 < (n+1)/(2n+1) < 1
lim(n->∞) n/2 < ∑(n:0->∞) (n+1)/(2n+1) < lim(n->∞) n

=>
∑(n:0->∞) (n+1)/(2n+1) -> ∞