a[n] = (n+1)/3^n
a[n+1]/a[n] = (n+2)/(3(n+1))
ρ = lim[a[n+1]/a[n]] = 1/3
所以收敛半径R = 1/ρ = 3
收敛区域 (-3,3)
回答楼上 hmlhmm
当x = -3的时候
(n+1)x^n/3^n = (n+1)(-1)^n
确实是交错级数,但是“一般项(?)”|(n+1)(-1)^n| = n+1趋近于∞,所以在-3处不收敛!