求1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n的最小值

求1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n的最小值
求1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n的最小值

求1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n的最小值
1/2n≤1/(1+n)＜1/n
1/2n≤1/(2+n)＜1/n
1/2n≤1/(3+n)＜1/n
1/2n≤1/(4+n)＜1/n
.
1/2n≤1/(n+n)＜1/n
上述式子相加
1/2<=1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n<1
最小值1/2

令f(n)=1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n
则f(n+1)=1/(2+n)+1/(3+n)+...+1/2n +1/(2n+1) +1/2(n+1)
f(n+1) - f(n) =1/(2n+1) +1/2(n+1) - 1/(1+n)
=1/(2n+1) - 1/(2n+2) >0
即f(n+1)>f(n)
所以f(n)是单调递增的
所以最小值f(1)=1/2

设s(n)=1/(1+n)+1/(2+n)+1/(3+n)+...+1/2n
s(n+1)=1/(1+n+1)+1/(2+n+1)+1/(3+n+1)+...+1/2(n+1)
=s(n)-1/(1+n)+1/(2n+1)+1/(2n+2)
令s(n+1)-s(n)=-1/(1+n)+1/(2n+1)+1/(2n+2)<0
满足这样的n使得s(n)减小。但上式恒不成立，也就是有s(n+1)>s(n)
那么s(n)的最小值在n=1得到，s(1)=1/2