用数学归纳法证明:1+1/2+1/3+…+1/(2^n-1)≤n 5
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当n=1时,左边=1,右边=1,成立。
假设你n=k时成立。即1+1/2+1/3+.....+1/(2^k-1)≤k
当n=k+1时,左边=1+1/2+1/3+.....+1/(2^k-1)+1/[(2^k-1)+1]+1/[(2^k-1)+2]
+1/[(2^k-1)+3]+……+1/[2^(k+1)-1]=1+1/2+1/3+.....+1/(2^k-1)+1/(2^k)+1/(2^k+1)
+1/(2^k+2)+……+1/[(2^k-1)+2^k]≤k+1/(2^k)+1/(2^k+1)+1/(2^k+2)+……+1/[2^(k+1)-1]
≤k+2^k*(1/2^k)
=k+1
综上可知:1+1/2+1/3+…+1/(2^n-1)≤n
(1/(2^k-1)+1/(2^k)+1/(2^k+1)+1/(2^k+2)+……+1/[(2^k-1)+2^k]是n=k+1添加的项,对应相等,且共2^k项)1/(2^k)+1/(2^k+1)+1/(2^k+2)+……+1/[(2^k-1)+2^k]去掉括号内+的1,2,3….. 2^k 后,则它们的和小于1
假设你n=k时成立。即1+1/2+1/3+.....+1/(2^k-1)≤k
当n=k+1时,左边=1+1/2+1/3+.....+1/(2^k-1)+1/[(2^k-1)+1]+1/[(2^k-1)+2]
+1/[(2^k-1)+3]+……+1/[2^(k+1)-1]=1+1/2+1/3+.....+1/(2^k-1)+1/(2^k)+1/(2^k+1)
+1/(2^k+2)+……+1/[(2^k-1)+2^k]≤k+1/(2^k)+1/(2^k+1)+1/(2^k+2)+……+1/[2^(k+1)-1]
≤k+2^k*(1/2^k)
=k+1
综上可知:1+1/2+1/3+…+1/(2^n-1)≤n
(1/(2^k-1)+1/(2^k)+1/(2^k+1)+1/(2^k+2)+……+1/[(2^k-1)+2^k]是n=k+1添加的项,对应相等,且共2^k项)1/(2^k)+1/(2^k+1)+1/(2^k+2)+……+1/[(2^k-1)+2^k]去掉括号内+的1,2,3….. 2^k 后,则它们的和小于1
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