ρ=lim(n->∞)|[1/(n+1)]/(1/n)|=lim(n->∞)|n/(1+n)|=1 收敛半径是R=1/ρ=1
当x=1时 ∑[x^(n+1)]/n=∑1/n 级数发散
当x=-1时 ∑[x^(n+1)]/n=∑[(-1)^(n+1)/n] 级数收敛
所以幂级数∑x^(n+1)/n的收敛区间是[-1,1)
令S(x)=∑x^(n+1)/n=x∑(x^n)/n=-xln(1-x) (-1