Question

求级数（-1）的n-1次幂*2的n+1次幂分之（2n+1）²是绝对收敛还是条件收敛

Answer 1

∑<n=0,∞>(-1)^(n-1) (2n+1)^2/2^(n+1)
ρ = lim<n→∞>|a<n+1>/a<n>|
= lim<n→∞>2^(n+1) (2n+3)^2/[2^(n+2)(2n+1)]
= lim<n→∞>(2n+3)^2/[2(2n+1)] = 1/2 < 1
原交错级数绝对收敛。