Question

求解高等数学敛散性，判断函数敛散性

Answer 1

ρ = lim<n→∞>a<n+1>/a<n>
= lim<n→∞>3*5...(2n+1)(2n+3) 4*7...(3n+1)/[4*7...(3n+1)(3n+4) 3*5...(2n+1)]
= lim<n→∞>(2n+3)/(3n+4) = lim<n→∞>(2+3/n)/(3+4/n) = 2/3 < 1
级数收敛。