1）级数的通项为
u(n) = (1/n)[(3/2)^n],
因
|u(n+1)/u(n)|
= [1/(n+1)][(3/2)^(n+1)]/(1/n)[(3/2)^n]
= (3/2)[n/(n+1)]
→ 3/2 > 1 (n→∞),
据比值判别法知原级数发散.
2）级数的通项为
u(n) = n[(3/4)^n],
因
|u(n+1)/u(n)|
= (n+1)[(3/4)^(n+1)]/n[(3/4)^n]
= (3/4)[(n+1)/n]
→ 3/4 < 1 (n→∞),
据比值判别法知原级数收敛.