不等式1/2+1/2等于多少?

证明不等式：1+1/2+1/3+…+1/(2&n)小于等于1/2+n.注：2&n是2的n次方

证明不等式：1+1/2+1/3+…+1/(2^n)≤1/2+n
①当n=1时,1+n/2=1.5,1+1/2+1/3+……+1/2^n=1.5,1/2+n=1.5,该式成立.
②假设n=k时该式成立,可得：(1+1/2+1/3+.+1/2^k)-(1+k/2)≥0,(1/2+k)-(1+1/2+1/3+.+1/2^k)≥0.
求证n=k+1时该式也成立.
③n=k+1时,原式的三项分别为1+(k+1)/2=1+k/2+1/2；1+1/2+1/3+.+1/2^(k+1)=1+1/2+1/3+.+1/2^k+1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)；1/2+k+1
两个≥号分别证明.
第一个≥号的证明：
1+1/2+1/3+.+1/2^k+1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)-(1+k/2+1/2)=(1+1/2+1/3+.+1/2^k-(1+k/2))+(1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)-1/2)
前一个括号里自然≥0；只要算后一个括号里即可：
显然,1/(2^k+1)>1/(2^k+2)>1/(2^k+3)>……>1/(2^k+2^k),且共有2^k项,所以1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)>1/(2^k+2^k)*2^k,即1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)>1/2,则第二个括号里>0,加起来自然≥0.
第二个≥号的证明：
1/2+k+1-(1+1/2+1/3+.+1/2^(k+1)=1+1/2+1/3+.+1/2^k+1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k))=(1/2+k-(1+1/2+1/3+.+1/2^(k+1)=1+1/2+1/3+.+1/2^k))+(1-(1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)))
同上,第一个括号里自然≥0；只要算后一个括号里即可：显然,1/(2^k+1)>1/(2^k+2)>1/(2^k+3)>……>1/(2^k+2^k),且共有2^k项,所以1/(2^k+1)+1/(2^k+2)+1/(2^k+3)+……+1/(2^k+2^k)

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