通项1/k=(1/k)lne>(1/k)ln(1+1/k)^k=ln(k+1)-lnk
∴[1+1/2+1/3+……+1/n+1/(n+1)](n->∞)
>[ln2-ln1+ln3-ln2+……+ln(n+2)-ln(n+1)](n->∞)
=ln(n+2)(n->∞)=∞
通项1/k=(1/k)lne>(1/k)ln(1+1/k)^k=ln(k+1)-lnk
∴[1+1/2+1/3+……+1/n+1/(n+1)](n->∞)
>[ln2-ln1+ln3-ln2+……+ln(n+2)-ln(n+1)](n->∞)
=ln(n+2)(n->∞)=∞