a(n+1)-an
=1/[(n+1)+1]+1/[(n+1)+2]+...+1/[(n+1)+(n-1)]+1/[(n+1)+n]+1/[(n+1)+(n+1)]
-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)-[1/(n+1)+1/(n+2)+...+1/(2n)]
=1/(2n+1)+1/(2n+2) -1/(n+1)
=1/(2n+1)-1/(2n+2)>0
a(n+1)>an，数列是递增数列。
2.
证：
n=1时，a1=1/(1+1)=1/2，不等式成立。
由第1问得数列是递增数列，即n≥2时，an>a1 an>1/2，不等式成立。
综上，得an≥1/2对一切正整数n恒成立。
设an=1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/2n(n属于正整数)则an+1-an=--
=1\/[(n+1)+1]+1\/[(n+1)+2]+...+1\/[(n+1)+(n-1)]+1\/[(n+1)+n]+1\/[(n+1)+(n+1)]-[1\/(n+1)+1\/(n+2)+...+1\/(2n)]=1\/(n+2)+1\/(n+3)+...+1\/(2n)+1\/(2n+1)+1\/(2n+2)-[1\/(n+1)+1\/(n+2)+...+1\/(2n)]=1\/(2n+1)+1\/(2n+2)-...
设An=1\/n+1+1\/n+2+1\/n﹢3+……+\/2n﹙n属于正整数)那么An+1-An=?
An=1\/n+1＋1／n+2＋1／n﹢3＋……＋／2nAn+1=1\/n+2+1\/n+3+...+1\/2n+1\/2n+1+1\/2n+2故An+1－An＝1\/2n+1+1\/2n+2-1\/n+1 =1\/2n+1-1\/2n+2 =1\/[（2n+1）（2n+2）]
设An=1\/n+1+1\/n+2+1\/n﹢3+……+\/2n﹙n属于正整数)那么An+1-An=?
An=1\/n+1＋1／n+2＋1／n﹢3＋……＋／2nAn+1=1\/n+2+1\/n+3+...+1\/2n+1\/2n+1+1\/2n+2故An+1－An＝1\/2n+1+1\/2n+2-1\/n+1 =1\/2n+1-1\/2n+2 =1\/[（2n+1）（2n+2）]
若数列{an}中的an=1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/2n(n∈N*)。
=an+ 1\/(2n+1)+1\/(2n+2)-1\/(n+1)=an+1\/(2n+1)-1\/(2n+2)选C
1\/(n+1)+1\/(n+2)+1\/(n+3)+ ……+1\/3n 极限
用定积分解答如下:
1.使不等式1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/(2n+1)
n=1：1\/2+1\/3 n=2：1\/3+1\/4+1\/5 n=3：1\/4+1\/5+1\/6+1\/7 n=4：1\/5+1\/6+1\/7+1\/8+1\/9,.看得出来,每次都是前面少一项,后面多两项,所以若令 an=1\/(n+1)+1\/(n+2)+1\/(n+3)+.+1\/(2n)+1\/(2n+1) ,那么 a(n+1)-an=1\/(2n+2)+1\/(2n+3)-1\/(n+1)...
设Sn=1\/(n+1)+1\/(n+2)+1\/(n+3)...1\/2n,求Sn的取值范围
详细解答如下，点击可放大：补充题：若n∈N,n≥2,求证:1\/2-1\/（n+1）<1\/2^2+1\/3^2+…+1\/n^2<1-1\/n 证明：对通项放缩：1\/n-1\/(n+1)=1\/[n(n+1)]＜1\/n^2＜1\/[n(n-1)]=1\/(n-1)-1\/n ∴原式＞(1\/2-1\/3)+(1\/3+1\/4)+…+[1\/n-1\/(n+1)]=1\/2-1\/...
1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/2n当n趋于无穷时的极限
1、本题的解答方法是：化无穷项的极限计算，成为定积分计算；2、若有疑问，欢迎追问，有问必答；若满意，请采纳。谢谢！3、每张图片均可点击放大。
设an=1+1\/2+1\/3+1\/4+...+1\/3n-1(n∈N*),则a(n+1)-an=
a(n+1) = 1+1\/2+1\/3+...+1\/(3n-1)+...+1\/[3(n+1)-1]= 1+1\/2+1\/3+...+1\/(3n-1)+...+1\/(3n+2)= 1+1\/2+1\/3+...+1\/(3n-1)+1\/3n+1\/(3n+1)+1\/(3n+2)an = 1+1\/2+1\/3+...+1\/(3n-1)a(n+1) - an = 1\/3n + 1\/(3n+1) + 1\/(...
利用放缩法证明1\/(n+1)+1\/(n+2)+1\/(n+3)+...+1\/(2n+1)<4\/5
放缩法证明方法：
