① When n= 1, 1+N/2 = 1.5,1+1/2+1/3+...

② Assuming n=k, this formula holds: (1+1/2+1/3+. + 1/2 k)-( 1+k/2) ≥ 0，(65438)

It is proved that the formula is also valid when n=k+ 1

③ When n = k+ 1, the three terms in the original formula are1+(k+1)/2 =1+k/2+1/2 respectively; 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

Two ≥ signs are proved respectively.

Proof of the first ≥ number:

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)

Natural ≥ 0 in the previous bracket; Just put it in brackets after calculation:

Obviously,1(2k+1) >1/(2 k+2) > 1/(2^k+3)>； ……& gt； 1/(2 k+2 k), * * has 2 k terms, so1/(2k+1)+1(2k+2)+1/. 1/(2 k+2 k) * 2 k, that is,1/(2k+1)+1(2k+2)+1/(2k+)1/. 0, add up to natural ≥0.

Proof of the second ≥ number:

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)+……+ / kloc-0//(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)))

Same as above, natural ≥ 0 in the first bracket; Just put it in the last bracket: obviously,1(2k+1) >1/(2 k+2) > 1/(2^k+3)>； ……& gt； 1/(2 k+2 k), * * has 2 k terms, so1/(2k+1)+1(2k+2)+1/.

The second question:

It is proved that when n= 1, (1+x)1=1+x;

Assuming n=k, the inequality holds, that is, (1+x) k > = 1+kx. Then when n=k+ 1,

( 1+x)^(k+ 1)=[( 1+x)^k]( 1+x)>； =( 1+kx)( 1+x)= 1+(k+ 1)x+kx^2>； 1+(k+ 1)x

Therefore, for n=k+ 1, the inequality holds.

And the principle of induction and the proof of inequality.