求极限lim n→∞(1/(n+1)+1/(n+2)+.+1/(n+n) 求极限(1/(n+1)+1/(n+2)+.+1/(n+n)

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求极限lim n→∞(1/(n+1)+1/(n+2)+.+1/(n+n) 求极限(1/(n+1)+1/(n+2)+.+1/(n+n)
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求极限lim n→∞(1/(n+1)+1/(n+2)+.+1/(n+n) 求极限(1/(n+1)+1/(n+2)+.+1/(n+n)
求极限lim n→∞(1/(n+1)+1/(n+2)+.+1/(n+n) 求极限(1/(n+1)+1/(n+2)+.+1/(n+n)

求极限lim n→∞(1/(n+1)+1/(n+2)+.+1/(n+n) 求极限(1/(n+1)+1/(n+2)+.+1/(n+n)
函数f(x)=1/(1+x).
用分点将区间[0,1]平均分成n份,分点是
x[k]=k/n,k=1,2,...,n.
利用定积分的定义,和式
∑{f(x[k])*(1/n),k=1...n}
当n->∞时的极限等于定积分
∫{f(x)dx,[0,1]}
而f(x[k])*(1/n)＝1/(n+k),通项相等,也就是说你的式子等于上面的和式.
于是
lim[1/(n+1) +1/(n+2)+1/(n+3)+……1/(n+n),n->∞]
=∫{f(x)dx,[0,1]}
=∫{1/(1+x)dx,[0,1]}
=ln(1+x)|[0,1]
=ln(1+1)-ln(1+0)
=ln2

答：
利用调和级数欧拉常数表达式：
1+1/2+1/3+1/4+...1/n = ln[n+1]+r[欧拉常数]
1/(n+1)+1/(n+2)+......+1/(n+n)
=1+1/2+1/3+……+1/n+1/(n+1)+1/(n+2)+......+1/(n+n)-(1+1/2+1/3+……+1/n)
=∑1/(n+n)-∑1/n
=ln[2...

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答：
利用调和级数欧拉常数表达式：
1+1/2+1/3+1/4+...1/n = ln[n+1]+r[欧拉常数]
1/(n+1)+1/(n+2)+......+1/(n+n)
=1+1/2+1/3+……+1/n+1/(n+1)+1/(n+2)+......+1/(n+n)-(1+1/2+1/3+……+1/n)
=∑1/(n+n)-∑1/n
=ln[2n+1]-ln[n+1]
=ln[(2n+1)/(n+1)]
所以：
原式=lim(n→∞)ln[(2n+1)/(n+1)]=ln[2]

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