已知数列｛an｝中,an=1+1/2+1/3+...+1/n,记sn=a1+a2+...+an用数学归纳法证明sn=(n+1)an-n

(i)当n=1时，左边=a1=1，右边=(1+1)a1-1=1 等式成立
(ii)假设当n=k(k属于N*,k>=1)时等式成立
即 sk=(k+1)ak-k
当n=k+1时 ak+1=ak+1/(1+k)
sk+1=sk+ak+1=(k+1)ak-k+ak+1
=(k+1+1)ak+1-k-(k+1)/(k+1)
=(k+1+1)ak+1-k-1
等式也成立
根据(i)(ii)可以断定sn=(n+1)an-n对于n属于N*都成立

(1) 当n=1时
S1=(1+1)a1-1
=2a1-1
=2-1
=1
∴当n=1时sn=(n+1)an-n成立
(2)假设n=k时sn=(n+1)an-n成立
即有 Sk=(k+1)ak-k
Sk+1=1+1/2+1/3+...+1/k+1/(k+1)
=Sk+1/(k+1)
=(k+1)ak-k+1/(k+1)
=(k+1)/k-k+1/(k+1)
=1+1/k-k+1/(k+1)
=(k+2)/(k+1)-(k+1)
=(k+2)ak+1-(k+1)
即当n=k+1时 Sk+1=(k+2)ak+1-(k+1) 成立
综上所述,当n∈N* 时有sn=(n+1)an-n 成立