微积分级数问题

Question

如题所述

Answer 1

5. 为缺项级数，记收敛半径为R，则
R^2 = lim<n→∞>a<n>/a<n+1>
= lim<n→∞>(2n)![(n+1)!]^2 / {[2(n+1)]!(n!)^2}
= lim<n→∞>(n+1)^2 / [(2n+2)(2n+1)]
= lim<n→∞>(n+1) / [2(2n+1)] = 1/4,
收敛半径 R = 1/2
x = ±1/2 时，级数发散。收敛域是 (-1/2, 1/2)