Question

数学归纳法

已知Sn=1+1/2+1/3+……+1/n(n大于1,n属于正整数)，用数学归纳法求证S（2^n）大于1+n/2

Answer 1

证明：
当n=2时，S(4)=1+1/2+1/3+1/4=1+13/12>1+1成立
假设n=k时，S(2^k)=1+1/2+1/3+...+1/2^k>1+k/2成立
当n=k+1时，S(2^(k+1))=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)+...+1/2^(k+1)
>1+k/2+2^k/2^(k+1)=1+k/2+1/2=1+(k+1)/2成立
所以S（2^n）>1+n/2

Answer 2

Let P(n) be the proposition.
Consider P(2)
L.H.S=1+1/2+1/3+1/4=25/12
R.H.S=1+1=2
As L.H.S.>R.H.S.
P(1) is true.
For all k belong to N\(1), assume that P(k) is true, i.e. 1+1/2+......+1/(2^k)>1+k/2

Consider for n=k+1

L.H.S.=1+1/2+1/3+.....+1/(2^k)+1/(2^k+1)+1/(2^k+2)+......1/(2^k+2^k)
(Note that 1/(2^k+2^k)=1/(2^(k+1))
>1+k/2+1/(2^k+1)+1/(2^k+2)+......1/(2^k+2^k) (by assumption)
>1+k/2+(2^k)(1/2)^(k+1) (As 1/(2^k+1), 1/(2^k+2),......., 1/(2^k+2^k)<(1/2)^(k+1)
(分母越大，数值越小)
=1+k/2+1/2
=1+(k+1)/2
=R.H.S.
So, by principle of M.I. P(n) is ture for all n belong to integers complement 1.

Answer 3

证明：
（1）当n=2时，S（2^2）=1+1/2+1/3+1/4=25/12>2
(2)假设当n=k时，S(2^K)>1+K/2
当n=k+1时，S(2^(K+1))=S(2^K)+1/(2^K+1)+...+1/(2^(K+1))>S(2^K)+2^K/(2^(K+1)) >1+K/2+1/2=1+(K+1)/2所以当n=k+1时，结论也成立
(3)所以结论得证即S（2^n）大于1+n/2恒成立(n大于1,n属于正整数)