(1)由 x>ln(1+x)(x≠ 0)得,1/n>ln(1+1/n). 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. |
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