f(n)=1+(1/2)+(1/3)+.(1/n) 求证:f(2的n次方)>(n+2)/2

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f(n)=1+(1/2)+(1/3)+.(1/n) 求证:f(2的n次方)>(n+2)/2

f(n)=1+(1/2)+(1/3)+.(1/n) 求证:f(2的n次方)>(n+2)/2
f(n)=1+(1/2)+(1/3)+.(1/n) 求证:f(2的n次方)>(n+2)/2

f(n)=1+(1/2)+(1/3)+.(1/n) 求证:f(2的n次方)>(n+2)/2
when n=2,we have,
f(4)=1+1/2+1/3+1/4=25/12>(2+2)/2=2
Assume when n=k,we have f(2^k)>(k+2)/2,then,
when n=k+1,f[2^(k+1)]=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^k +2^k) > (k+2)/2 + 2^k * 1/(2^k +2^k) = [(k+1)+2]/2
So when n>1,we have f(2^n)>(n+2)/2

1+(1/2)+(1/3)+.........(1/2^n) >(n/2)+1
(lnx)'=1/x
1+(1/2)+(1/3)+.........(1/2^n)
=1+[(1/2)+(1/3)+.........(1/2^n)]
>(从1到2^n)∫(1/x)
=ln(2^n)-ln1
=nln2
>1+n/2
n>=6时,一定成立
n=1,2,3,4,5容易验证成立。

数学归纳法:n=2时,左边=25/12,右边=2,左边>右边,成立
假设当n=k时有f(2^k)>(k+2)/2,则
n=k+1时,f[2^(k+1)]=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^k +2^k) > (k+2)/2 + 2^k * 1/(2^k +2^k) = [(k+1)+2]/2
所以由数学归纳法得当 n>1, 有f(2^n)>(n+2)/2

when n=2, we have,
f(4)=1+1/2+1/3+1/4=25/12>(2+2)/2=2
Assume when n=k, we have f(2^k)>(k+2)/2, then,
when n=k+1, f[2^(k+1)]=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^k +2^k) > (k+2)/2 + 2^k *...

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when n=2, we have,
f(4)=1+1/2+1/3+1/4=25/12>(2+2)/2=2
Assume when n=k, we have f(2^k)>(k+2)/2, then,
when n=k+1, f[2^(k+1)]=f(2^k)+1/(2^k+1)+1/(2^k+2)+...+1/(2^k +2^k) > (k+2)/2 + 2^k * 1/(2^k +2^k) = [(k+1)+2]/2
So when n>1, we have f(2^n)>(n+2)/2
1+(1/2)+(1/3)+.........(1/2^n) >(n/2)+1
(lnx)'=1/x
1+(1/2)+(1/3)+.........(1/2^n)
=1+[(1/2)+(1/3)+.........(1/2^n)]
>(从1到2^n)∫(1/x)
=ln(2^n)-ln1
=nln2
>1+n/2
n>=6时,一定成立
n=1,2,3,4,5容易验证成立。

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数学归纳法