Question

已知：Sn=1+1/2+1/3+……+1/n,用数学归纳法证明：Sn^2>1+n/2(n>=2,n∈N+）

Answer 1

Sn=1/1*2+1/2*3,...,1/n*(n+1)
=（1-1/2）+（1/2-1/3）+......+[1/n-1/(n+1)]
=1-1/(n+1)
=n/(n+1)
用数学归纳法证:
当k=1时:S1=1/1*2=1/2
k/(k+1)=1/2
所以Sk=k/(k+1)
假设当k=n时成立,即:Sn=n/(n+1)
那么当k=n+1时,证明S(n+1)=(n+1)/(n+2)即可
S(n+1)=1/1*2+1/2*3,...,1/n*(n+1)+1/(n+1)(n+2)
=n/(n+1)+1/(n+1)(n+2)
=n(n+2)/(n+1)(n+2)+1/(n+1)(n+2)
=(n^2+2n+1)/(n+1)(n+2)
=(n+1)^2/(n+1)(n+2)
=(n+1)/(n+2)
所以综上:Sn=n/(n+1)

Answer 2

1+n/2≤1+1/2+1/3+...+1/(2^n)≤1/2+n
证明：
(1)当n=1时，1+1/2<=1+1/2<=1/2+1
，原不等式成立。
(2)设n=k时，命题成立。即有：
1+k/2≤1+1/2+1/3+...+1/(2^k)≤1/2+k
（3）当n=k+1时，
1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)
>=1+k/2+1/(2^k+1)+...+1/2^(k+1)
>1+k/2+1/2^(k+1)+...+1/2^(k+1)
>1+k/2+[2^(k+1)-2^k]/2^(k+1)=1+(k+1)/2

1+1/2+1/3+...+1/2^k+1/(2^k+1)+...+1/2^(k+1)
<=1/2+k+1/(2^k+1)+...+1/2^(k+1)
<1+k+1/2^k+...+1/2^k
<1+k+[2^(k+1)-2^k]/2^k=1+(k+1)
即n=k+1时，原不等式成立。
故原命题成立。