Question

∑1/ln(n+1)敛散性

Answer 1

正项级数,用比值审敛法：
lim（n→∞）u(n+1)/un=[1/ln(n+2)]/[1/ln(n+1)]
=lim（n→∞）ln(n+1)/ln(n+2) ＜1,级数收敛