Question

1+1/√2+1/(√3)^3+……+1/(√n)^n 判断其收敛性

Answer 1

a = 1/(√n)^n ,

ρ= lima/a

= lim(√n)^n / [√(n+1)]^(n+1)

= lim{√[n/(n+1)]}^n / √(n+1) = 0

故级数收敛。