令 x=-1/(1+y) 得,ln(1+1/y)>1/(y+1).(2)Sn=1+1/2+1/3+……+1/n
>ln(1+1)+ln(1+1/2)+ln(1+1/3)+……+ln(1+1/n)
=ln(n+1)
故(n→∞)Sn>(n→∞)ln(n+1)=∞(发散)
Sn=1+1/2+1/3+……+1/n
<1+ln(1+1)+ln(1+1/2)+……+ln[1+/(n-1)]
=1+ln(n)
(3)Pn=(n→∞)[(1+1/2+1/3+……+1/n-ln(n)]
>(n→∞)[ln(1+1/n)]=0(有下界)
(4)Pn=(n→∞)[(1+1/2+1/3+……+1/n-ln(n)<1
(5)Pn-Pn+1=ln(1+1/n)-1/(n+1)>0.(单调递减)
(6)由单调有界数列极限定理得(n→∞)Pn=欧拉常数c.